Engineering Mathematics I - Unit 1 Notes PDF

For more Subjects https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐔𝐧𝐢𝐭 𝐈 ∶ 𝐃𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐥 𝐂𝐚𝐥𝐜𝐮𝐥𝐮𝐬 𝐑𝐨𝐥𝐥𝐞’𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: Let f(x) be a function defined in [a, b] i) function f(x) is continuous on the closed interval [a, b] ii) differentiable...

For more Subjects https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐔𝐧𝐢𝐭 𝐈 ∶ 𝐃𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐭𝐢𝐚𝐥 𝐂𝐚𝐥𝐜𝐮𝐥𝐮𝐬 𝐑𝐨𝐥𝐥𝐞’𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: Let f(x) be a function defined in [a, b] i) function f(x) is continuous on the closed interval [a, b] ii) differentiable on the open interval (a, b) iii) f(a) = f(b) then there exists at least one point 𝑥 = 𝑐 in the open interval (a, b) such that f′(c) = 0. 𝐆𝐞𝐨𝐦𝐞𝐭𝐫𝐢𝐜 𝐢𝐧𝐭𝐞𝐫𝐩𝐫𝐞𝐭𝐚𝐭𝐢𝐨𝐧 There is a point c on the interval (𝑎, 𝑏) where the tangent to the graph of the function is horizontal. 𝐋𝐚𝐠𝐫𝐚𝐧𝐠𝐞’𝐬 𝐌𝐞𝐚𝐧 𝐕𝐚𝐥𝐮𝐞 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: Let f(x) be a function defined in [a, b] i) Function f(x) is continuous on a closed interval [a , b] ii) Function f(x) differentiable on the open interval (a, b) then there is at least one point x = c on this interval (a, b), such that 𝑓(𝑏) − 𝑓(𝑎) 𝑓 ′ (𝑐) = 𝑏−𝑎 𝐆𝐞𝐨𝐦𝐞𝐭𝐫𝐢𝐜 𝐈𝐧𝐭𝐞𝐫𝐩𝐫𝐞𝐭𝐚𝐭𝐢𝐨𝐧 The chord passing through the points of the graph corresponding to the ends of the segment a and b has the slope equal to 𝑘= 𝑓(𝑏) − 𝑓(𝑎) tan 𝛼 = 𝑏−𝑎 Then there is a point 𝑥 = 𝑐 inside the interval [a , b] where the tangent to the graph is parallel to the chord. "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐂𝐚𝐮𝐜𝐡𝐲’𝐬 𝐌𝐞𝐚𝐧 𝐕𝐚𝐥𝐮𝐞 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: Let f(x) be a function defined in [a, b] i) Function f(x) and g(𝑥) is continuous on a closed interval [a , b] ii) Function f(x) and g(𝑥) differentiable on the open interval (a, b) iii) g ′ (𝑥) ≠ 0 for all value of x in (a, b) then there is at least one point x = c on this interval (a, b), such that 𝑓 ′ (𝑐) 𝑓(𝑏) − 𝑓(𝑎) = g ′ (𝑐) g(𝑏) − g(𝑎) *********** Example 1: Verify Rolle’s Mean Value theorem for 𝒇(𝒙) = 𝒙𝟐 − 𝟓𝒙 + 𝟒 𝑖𝑛 [𝟏, 𝟒] Solution : 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 4 𝐴𝑠 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 4 is a polynomial Every polynomial is continuous and differentiable everywhere ∴ 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 4 is continuous in [1 , 4] and differetiable in (1, 4) 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 4 𝑓(𝑎) = 𝑓(1) = 12 − 5(1) + 4 = 1 − 5 + 4 = 0 𝑓(𝑏) = 𝑓(4) = 42 − 5(4) + 4 = 16 − 20 + 4 = 0 𝑓(𝑎) = 𝑓(𝑏) All condition of Rolle’s Mean Value theorem satisfied. then their exist at least one c ∈ (1, 4) such that 𝑓 ′ (𝑐) = 0 𝑓(𝑥) = 𝑥 2 − 5𝑥 + 4 𝑓 ′ (𝑥) = 2𝑥 − 5 put 𝑥 = 𝑐 𝑓 ′ (𝑐) = 2𝑐 − 5 5 ∴ 𝑓 ′ (𝑐) = 0 ⇒ 2𝑐 − 5 = 0 ∴ 2𝑐 = 5 ∴ 𝑐 = = 2.5 2 ∴ 𝑐 = 2.5 ∈ (1, 4) hence Lagrange’s Mean Value theorem verified. ********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example 2: Verify Rolle’s Mean Value theorem for 𝜋 5𝜋 𝑓(𝑥) = 𝑒 𝑥 (𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠 𝑥) 𝑖𝑛 [ , ] 4 4 Solution : 𝑓(𝑥) = 𝑒 𝑥 (𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠 𝑥) 𝐴𝑠 𝑓(𝑥) is a combination of exponational and sine, cosine functions Exponational and Sine, Cosine function are continuous and differentiable 𝜋 5𝜋 𝜋 5𝜋 ∴ 𝑓(𝑥) = 𝑒 𝑥 (𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠 𝑥) is continuous in [ , ] and differetiable in ( , ) 4 4 4 4 𝑓(𝑥) = 𝑒 𝑥 (𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠 𝑥) = 𝜋 𝜋 𝜋 𝜋 𝜋 1 1 𝑓(𝑎) = 𝑓 ( ) = 𝑒 4 (𝑠𝑖𝑛 − 𝑐𝑜𝑠 ) = 𝑒 4 ( − )=0 4 4 4 √2 √2 5𝜋 5𝜋 5𝜋 5𝜋 5𝜋 1 1 𝑓(𝑏) = 𝑓 ( ) = 𝑒 4 (𝑠𝑖𝑛 − 𝑐𝑜𝑠 )=𝑒4 (− − (− )) = 0 4 4 4 √2 √2 𝑓(𝑎) = 𝑓(𝑏) All condition of Rolle’s Mean Value theorem satisfied. 𝜋 5𝜋 then their exist at least one c ∈ ( , ) such that 𝑓 ′ (𝑐) = 0 4 4 𝑓(𝑥) = 𝑒 𝑥 (𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠 𝑥) 𝑓 ′ (𝑥) = 𝑒 𝑥 (𝑐𝑜𝑠 𝑥 + 𝑠𝑖𝑛 𝑥) + 𝑒 𝑥 (𝑠𝑖𝑛 𝑥 − 𝑐𝑜𝑠 𝑥) 𝑓 ′ (𝑥) = 2𝑒 𝑥 sin 𝑥 put 𝑥 = 𝑐 𝑓 ′ (𝑐) = 2𝑒 𝑐 sin 𝑐 ∴ 𝑓 ′ (𝑐) = 0 ⇒ 2𝑒 𝑐 sin 𝑐 = 0 ∴ sin 𝑐 = 0 ∴ 𝑐 = sin−1 0 ∴ 𝑐 = 𝑛𝜋 𝑛 = 0,1,2,3 …. 𝑐 = 0, 𝜋, 2𝜋, 3𝜋, 4𝜋 … 𝜋 5𝜋 ∴𝑐=𝜋 ∈( , ) hence Rolle’s Mean Value theorem verified. 4 4 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example 3: Verify Lagrange’s Mean Value theorem for 𝒇(𝒙) = (𝒙 − 𝟏)(𝒙 − 𝟐)(𝒙 − 𝟑) 𝑖𝑛 [𝟎 𝟑] Solution : 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3) 𝑓(𝑥) = 𝑥 3 − 6𝑥 2 + 11𝑥 − 6 𝐴𝑠 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3) is a polynomial Every polynomial is continuous and differentiable everywhere ∴ 𝑓(𝑥) = (𝑥 − 1)(𝑥 − 2)(𝑥 − 3) is continuous in [0, 3] and differetiable in (0 3) All condition of Lagrange’s Mean Value theorem satisfied. 𝑓(𝑏)− 𝑓(𝑎) then their exist at least one c ∈ (𝑎, 𝑏) such that 𝑓 ′ (𝑐) = 𝑏−𝑎 𝑓(𝑥) = 𝑥 3 − 6𝑥 2 + 11𝑥 − 6 𝑓(𝑎) = 𝑓(0) = (0)3 − 6(0)2 + 11(0) − 6 = −6 𝑓(𝑏) = 𝑓(3) = (3)3 − 6(3)2 + 11(3) − 6 = 27 − 54 + 33 − 6 = 0 𝑓 ′ (𝑥) = 3𝑥 2 − 12𝑥 + 11 put 𝑥 = 𝑐 𝑓 ′ (𝑐) = 3𝑐 2 − 12𝑐 + 11 𝑓(𝑏)− 𝑓(𝑎) 𝑓 ′ (𝑐) = 𝑏−𝑎 0−(−6) 6 ∴ 3𝑐 2 − 12𝑐 + 11 = = 3−0 3 ∴ 3𝑐 2 − 12𝑐 + 11 = 2 ∴ 3𝑐 2 − 12𝑐 + 11 − 2 = 0 ∴ 3𝑐 2 − 12𝑐 + 9 = 0 ∴ 𝑐 2 − 4𝑐 + 3 = 0 ∴ 𝑐 2 − 3𝑐 − 𝑐 + 3 = 0 ∴ 𝑐(𝑐 − 3) − 1(𝑐 − 3) = 0 ∴ (𝑐 − 3)(𝑐 − 1) = 0 ∴ 𝑐 = 3 ,1 ∴ 𝑐 = 1 ∈ (0, 3) hence Lagrange’s Mean Value theorem verified. "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example 4: Verify Lagrange’s Mean Value theorem for 𝒇(𝒙) = 𝒍𝒐𝒈 𝒙 𝑖𝑛 [𝟏, 𝒆] Solution : 𝑓(𝑥) = 𝑙𝑜𝑔 𝑥 𝐴𝑠 𝑓(𝑥) = 𝑙𝑜𝑔 𝑥 is a logarithmic function Every logarithmic function is continuous and differentiable in its domain ∴ 𝑓(𝑥) = 𝑙𝑜𝑔 𝑥 is continuous in [1, e] and differetiable in (1 e) All condition of Lagrange’s Mean Value theorem satisfied. 𝑓(𝑏)− 𝑓(𝑎) then their exist at least one c ∈ (1 e) such that 𝑓 ′ (𝑐) = 𝑏−𝑎 𝑓(𝑥) = 𝑙𝑜𝑔 𝑥 𝑓(𝑎) = 𝑓(0) = 𝑙𝑜𝑔 1 = 0 𝑓(𝑏) = 𝑓(𝑒) = 𝑙𝑜𝑔 𝑒 = 1 1 1 𝑓 ′ (𝑥) = put 𝑥 = 𝑐 𝑓 ′ (𝑐) = 𝑥 𝑐 1 1−0 = ∴𝑐 = 𝑒−1 𝑐 𝑒−1 ∴ 𝑐 = 𝑒 − 1 ∈ (1 e) hence Lagrange’s Mean Value theorem verified. *********** Example 5: Verify Cauchy’s Mean value theorem theorem for 𝒇(𝒙) = 𝒙𝟑 𝑎𝑛𝑑 𝐠(𝒙) = 𝒙𝟒 𝑖𝑛 [𝟎 𝟐] Solution : 𝑓(𝑥) = 𝑥 3 and g(𝑥) = 𝑥 4 𝐴𝑠 𝑓(𝑥) = 𝑥 3 and g(𝑥) = 𝑥 4 are a polynomials Every polynomial is continuous and differentiable everywhere ∴ 𝑓(𝑥) = 𝑥 3 and g(𝑥) = 𝑥 4 is continuous in [0, 2] and differetiable in (0 2) As g(𝑥) = 𝑥 4 g ′ (𝑥) = 4𝑥 3 ≠ 0 for 𝑥 in (0, 2) All condition of Cauchy’s Mean Value theorem satisfied. "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝑓′ (𝑐) 𝑓(𝑏)− 𝑓(𝑎) then their exist at least point c ∈ (0, 2) such that = g′ (c) g(𝑏)− g(𝑎) 𝐴𝑠 𝑓(𝑥) = 𝑥 3 and g(𝑥) = 𝑥 4 𝑓(𝑏) = 𝑓(2) = 23 = 8 and g(𝑏) = g(2) = 24 = 16 𝑓(𝑎) = 𝑓(0) = 0 and g(𝑎) = g(0) = 0 f ′ (𝑥) = 3𝑥 2 and g ′ (𝑥) = 4𝑥 3 Put x = c f ′ (𝑐) = 3𝑐 2 and g ′ (𝑐) = 4𝑐 3 3𝑐 2 8− 0 = 4𝑐 3 16 − 0 3 1 6 3 = ∴ 4c = 6 ∴ c= ∴ c= 4𝑐 2 4 2 3 ∴ c= ∈ (0, 2) hence Cauchy’s Mean value theorem verified. 2 *********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝐨𝐟 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧 𝐓𝐚𝐲𝐥𝐨𝐫 ′ 𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: Statement ∶ Let 𝑓(𝑎 + ℎ) be a function of h which can be expanded in powers of h and let the expansion be differentiable term by term any number of times w. r. t. h ℎ2 ℎ3 ℎ𝑛 then 𝑓(𝑎 + ℎ) = 𝑓(𝑎) + ℎ 𝑓 ′ (𝑎) + 𝑓 ′′ (𝑎) + 𝑓 ′′′ (𝑎) + … … + 𝑓 𝑛 (𝑎)+... 2! 3! 𝑛! 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝐨𝐟 𝐟(𝐱 + 𝐡) 𝐢𝐧 𝐩𝐨𝐰𝐞𝐫 𝐨𝐟 ′𝐡′ ℎ2 ℎ3 ℎ𝑛 𝑓(𝑥 + ℎ) = 𝑓(𝑥) + ℎ 𝑓 ′ (𝑥) + 𝑓 ′′ (𝑥) + 𝑓 ′′′ (𝑥) + … ….. + 𝑓 𝑛 (𝑥) + … …. 2! 3! 𝑛! 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝐨𝐟 𝐟(𝐱 + 𝐡) 𝐢𝐧 𝐩𝐨𝐰𝐞𝐫 𝐨𝐟 ′𝐱′ 𝑥2 𝑥3 𝑥𝑛 𝑓(𝑥 + ℎ) = 𝑓(ℎ) + 𝑥 𝑓 ′ (ℎ) + 𝑓 ′′ (ℎ) + 𝑓 ′′′ (ℎ) + … ….. + 𝑓 𝑛 (ℎ) + … …. 2! 3! 𝑛! 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝐟(𝐱) 𝐢𝐧 𝐩𝐨𝐰𝐞𝐫 𝐨𝐟 (𝐱 − 𝐚) = 𝟎 𝐨𝐫 𝐚𝐛𝐨𝐮𝐭 𝐱 = 𝐚 (𝑥 – 𝑎)2 (𝑥 – 𝑎)3 (𝑥 – 𝑎)𝑛 𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎) 𝑓 ′ (𝑎) + 𝑓 ′′ (𝑎) + 𝑓 ′′′ (𝑎) + … + 𝑓 𝑛 (𝑎) …. 2! 3! 𝑛! ******** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example 1: Using Taylor’s theorem express (x − 2)4 − 3(x − 2)3 + 4(x − 2)2 + 5 in powers of x Solution: Let f(x + h) = (x − 2)4 − 3(x − 2)3 + 4(x − 2)2 + 5 Here 𝑥 + ℎ = 𝑥 − 2 ∴ ℎ = −2 By using Taylor’s theorem, 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝐨𝐟 𝐟(𝐱 + 𝐡) 𝐢𝐧 𝐩𝐨𝐰𝐞𝐫 𝐨𝐟 𝐱 ′ (ℎ) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥 4 ′𝑣 𝑓(𝑥 + ℎ) = 𝑓(ℎ) + 𝑥 𝑓 + 𝑓 (ℎ) + 𝑓 (ℎ) + 𝑓 (ℎ) + … … … … 2! 3! 4! 𝑥2 𝑥3 𝑥4 𝑓(𝑥 − 2) = 𝑓(−2) + 𝑥 𝑓 ′ (−2) + 𝑓 ′′ (−2) + 𝑓 ′′′ (−2) + 𝑓 ′𝑣 (−2) + … … (𝐴) 2! 3! 4! ∴ 𝑓(𝑥) = 𝑥 4 − 3𝑥 3 + 4𝑥 2 + 5 𝑓(𝑥) = 𝑥 4 − 3𝑥 3 + 4𝑥 2 + 5 ∴ 𝑓(ℎ) = 𝑓(−2) = 61 𝑓 ′ (𝑥) = 4𝑥 3 − 9𝑥 2 + 8𝑥 ∴ 𝑓 ′ (ℎ) = 𝑓 ′ (−2) = −84 𝑓 ′′ (𝑥) = 12𝑥 2 − 18𝑥 + 8 ∴ 𝑓 ′′ (−2) = 𝑓 ′′ (−2) = 92 𝑓 ′′′ (𝑥) = 24𝑥 − 18 ∴ 𝑓 ′′′ (−2) = 𝑓 ′′′ (−2) = −66 𝑓 ′𝑣 (𝑥) = 24 ∴ 𝑓 ′𝑣 (−2) = 𝑓 ′𝑣 (−2) = 24 𝑓 𝑣 (𝑥) = 0 ∴ 𝑓 𝑣 (−2) = 𝑓 𝑣 (−2) = 0 Equation (A) becomes 𝑥2 𝑥3 𝑥4 𝑓(𝑥 − 2) = 61 + 𝑥 (−84) + (92) + (−66) + (24) 2 6 24 𝑓(𝑥 − 2) = 61 − 84𝑥 + 46𝑥 2 − 11𝑥 3 + 𝑥 4 ********* "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example 2: Using Taylor’s theorem express (𝑥 + 2)4 + 3(𝑥 + 2)3 + (𝑥 + 2) + 7 in powers of x Solution: Let 𝑓(𝑥 + ℎ) = (𝑥 + 2)4 + 3(𝑥 + 2)3 + (𝑥 + 2) + 7 Here 𝑥 + ℎ = 𝑥 + 2 ∴ℎ=2 By using Taylor’s theorem, 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝐨𝐟 𝐟(𝐱 + 𝐡) 𝐢𝐧 𝐩𝐨𝐰𝐞𝐫 𝐨𝐟 𝐱 ′ (ℎ) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥 4 ′𝑣 𝑓(𝑥 + ℎ) = 𝑓(ℎ) + 𝑥 𝑓 + 𝑓 (ℎ) + 𝑓 (ℎ) + 𝑓 (ℎ) + … … … … 2! 3! 4! 𝑥2 𝑥3 𝑥4 𝑓(𝑥 − 2) = 𝑓(2) + 𝑥 𝑓 ′ (2) + 𝑓 ′′ (2) + 𝑓 ′′′ (2) + 𝑓 ′𝑣 (2) + … … (𝐴) 2! 3! 4! ∴ 𝑓(𝑥) = 𝑥 4 + 3𝑥 3 + 𝑥 + 7 𝑓(𝑥) = 𝑥 4 + 3𝑥 3 + 𝑥 + 7 ∴ 𝑓(2) = 49 𝑓 ′ (𝑥) = 4𝑥 3 + 9𝑥 2 + 1 ∴ 𝑓 ′ (2) = 69 𝑓 ′′ (𝑥) = 12𝑥 2 + 18𝑥 ∴ 𝑓 ′′ (2) = 84 𝑓 ′′′ (𝑥) = 24𝑥 + 18 ∴ 𝑓 ′′′ (2) = 66 𝑓 ′𝑣 (𝑥) = 24 ∴ 𝑓 ′𝑣 (2) = 24 𝑓 𝑣 (𝑥) = 0 ∴ 𝑓 𝑣 (2) = 0 Equation (A) becomes 𝑥2 𝑥3 𝑥4 𝑓(𝑥 + 2) = 49 + 𝑥 (69) + (84) + (66) + (24) + 0 2 6 24 𝑓(𝑥 + 2) = 49 + 69𝑥 + 42𝑥 2 + 11𝑥 3 + 𝑥 4 ********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example 3: Using Taylor’s theorem express 49 + 69𝑥 + 42𝑥 2 + 11𝑥 3 + 𝑥 4 in powers of (𝑥 + 2) Solution: Let 𝑓(𝑥) = 49 + 69𝑥 + 42𝑥 2 + 11𝑥 3 + 𝑥 4 Here 𝑥−𝑎 =𝑥+2 ∴ −𝑎 = 2 ∴ 𝑎 = −2 By using Taylor’s theorem, 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝐟(𝐱) 𝐢𝐧 𝐩𝐨𝐰𝐞𝐫 𝐨𝐟 (𝐱 − 𝐚) 𝐨𝐫 𝐚𝐛𝐨𝐮𝐭 𝐱 = 𝐚 (𝑥 – 𝑎)2 (𝑥 – 𝑎)3 (𝑥 – 𝑎)𝑛 𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎) 𝑓 ′ (𝑎) + 𝑓 ′′ (𝑎) + 𝑓 ′′′ (𝑎) + … + 𝑓 𝑛 (𝑎) …. 2! 3! 𝑛! (𝑥 + 2)2 (𝑥 + 2)3 (𝑥 + 2)4 𝑓 (𝑥 ) = 𝑓(−2) + (𝑥 + 2)𝑓 ′ (−2) + 𝑓 ′′ (−2) + 𝑓 ′′′ (−2) + 𝑓 ′𝑣 (−2) …. (𝐴) 2 6 24 ∴ 𝑓(𝑥) = 49 + 69𝑥 + 42𝑥 2 + 11𝑥 3 + 𝑥 4 𝑓(𝑥) = 49 + 69𝑥 + 42𝑥 2 + 11𝑥 3 + 𝑥 4 ∴ 𝑓(−2) = 7 𝑓 ′ (𝑥) = 69 + 84𝑥 + 33𝑥 2 + 4𝑥 3 ∴ 𝑓 ′ (−2) = 1 𝑓 ′′ (𝑥) = 84 + 66𝑥 + 12𝑥 2 ∴ 𝑓 ′′ (−2) = 0 𝑓 ′′′ (𝑥) = 66 + 24𝑥 ∴ 𝑓 ′′′ (−2) = 18 𝑓 ′𝑣 (𝑥) = 24 ∴ 𝑓 ′𝑣 (−2) = 24 𝑓 𝑣 (𝑥) = 0 ∴ 𝑓 𝑣 (−2) = 0 Equation (A) becomes (𝑥 + 2)2 (𝑥+2)3 (𝑥+2)4 𝑓(𝑥 ) = 7 + (𝑥 + 2)1 + (0) + 18 + 24 2 6 24 𝑓(𝑥 ) = 7 + (𝑥 + 2) + 3(𝑥 + 2)3 + (𝑥 + 2)4 ********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example 4: Using Taylor’s theorem express 𝑥 3 + 7𝑥 2 + 𝑥 − 6 in powers of (𝑥 − 3) Solution: Let 𝑓(𝑥) = 𝑥 3 + 7𝑥 2 + 𝑥 − 6 Here 𝑥 − 𝑎 = 𝑥 − 3 ∴𝑎=3 By using Taylor’s theorem, 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧 𝐟(𝐱) 𝐢𝐧 𝐩𝐨𝐰𝐞𝐫 𝐨𝐟 (𝐱 − 𝐚) 𝐨𝐫 𝐚𝐛𝐨𝐮𝐭 𝐱 = 𝐚 (𝑥 – 𝑎)2 (𝑥 – 𝑎)3 (𝑥 – 𝑎)𝑛 𝑓(𝑥) = 𝑓(𝑎) + (𝑥 − 𝑎) 𝑓 ′ (𝑎) + 𝑓 ′′ (𝑎) + 𝑓 ′′′ (𝑎) + … + 𝑓 𝑛 (𝑎) …. 2! 3! 𝑛! (𝑥 – 3)2 (𝑥 – 3)3 (𝑥 – 3)4 𝑓(𝑥 ) = 𝑓(3) + (𝑥 − 3)𝑓 ′ (3) + 𝑓 ′′ (ℎ) + 𝑓 ′′′ (3) + 𝑓 ′𝑣 (3)+...(A) 2! 3! 4! ∴ 𝑓(𝑥) = 𝑥 3 + 7𝑥 2 + 𝑥 − 6 𝑓(𝑥) = 𝑥 3 + 7𝑥 2 + 𝑥 − 6 ∴ 𝑓(3) = 87 𝑓 ′ (𝑥) = 3𝑥 2 + 14𝑥 + 1 ∴ 𝑓 ′ (3) = 70 𝑓 ′′ (𝑥) = 6𝑥 + 14 ∴ 𝑓 ′′ (3) = 32 𝑓 ′′′ (𝑥) = 6 ∴ 𝑓 ′′′ (3) = 6 𝑓 ′𝑣 (𝑥) = 0 ∴ 𝑓 ′𝑣 (3) = 0 Equation (A) becomes (𝑥 − 3)2 (𝑥 − 3)3 𝑓(𝑥 ) = 87 + (𝑥 − 3)69 + 32 + 6 2 6 𝑓(𝑥 ) = 87 + 70(𝑥 − 3) + 16(𝑥 − 3)2 + (𝑥 − 3)3 ********** Example 5: Using Taylor’s theorem express (𝑥 − 1)4 − 3(𝑥 − 1)3 + 4(𝑥 − 1)2 + 5 in powers of x. Example 6: Using Taylor’s theorem express 2(𝑥 − 2)3 + 19(𝑥 − 2)2 + 53(𝑥 − 2) + 40 in powers of x. Example 7: Using Taylor’s theorem express 3𝑥 3 − 2𝑥 2 + 𝑥 − 6 in powers of 𝑥 − 2 Example 8: Using Taylor’s theorem express 1 + 2𝑥 + 3𝑥 2 + 4𝑥 3 in powers of 𝑥 + 1 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐌𝐚𝐜𝐥𝐚𝐮𝐫𝐢𝐧′ 𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: 𝐒𝐭𝐚𝐭𝐞𝐦𝐞𝐧𝐭: Let f(x) be a function of x which can be expanded in ascending powers and let the expansion be differentiable term by term any number of times then ′ (0) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 + 𝑓 (0) + 𝑓 (0) + … ….. + 𝑓 (0) + … …. 2! 3! 𝑛! 𝐍𝐨𝐭𝐞: 1) If y = f(x) then f(0) = (y)0 , f ′ (0) = (y1 )0 , f ′′ (0) = (y2 )0 … … f n (0) = (yn )0 Maclaurin′ s Theorem stated as x2 x3 xn y = (y)0 + x(y1 )0 + (y2 )0 + (y3 )0 + … ….. + (yn )0 + … …. 2! 3! n! xn n 2) The (n + 1)𝑡ℎ term of expansion f (0) is called general term. n! ex + e−x ex − e−x 3) cosh x = sinh x = 2 2 d 4) (cosh x) = sinh x dx d 5) (sinh x) = cosh x dx 6) ∫ sinh x 𝑑𝑥 = cosh 𝑥 + c 7) ∫ cosh x 𝑑𝑥 = sinh 𝑥 + c "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example : Expansion of ex Solution : 𝐿𝑒𝑡 𝑓(𝑥) = 𝑒 𝑥 by Maclaurin′ s theorem ′ (0) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 + 𝑓 (0) + 𝑓 (0) + ……..+ 𝑓 (0) + … …. 2! 3! 𝑛! 𝑓(𝑥) = 𝑒 𝑥 ∴ 𝑓(0) = 𝑒 0 = 1 𝑓 ′ (𝑥) = 𝑒 𝑥 ∴ 𝑓 ′ (0) = 𝑒 0 = 1 𝑓 ′′ (𝑥) = 𝑒 𝑥 ∴ 𝑓 ′′ (0) = 𝑒 0 = 1 𝑓 ′′′ (𝑥) = 𝑒 𝑥 ∴ 𝑓 ′′′ (0) = 𝑒 0 = 1 𝑓 ′𝑣 (𝑥) = 𝑒 𝑥 ∴ 𝑓 ′𝑣 (0) = 𝑒 0 = 1 𝑓 𝑣 (𝑥) = 𝑒 𝑥 ∴ 𝑓 𝑣 (0) = 𝑒 0 = 1 ………….. 𝑓 𝑛 (𝑥) = 𝑒 𝑥 ∴ 𝑓 𝑛 (0) = 𝑒 0 = 1 by Maclaurin′ s Theorem: ′ (0) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 + 𝑓 (0) + 𝑓 (0) + … ….. + 𝑓 (0) + … …. 2! 3! 𝑛! 𝑥2 𝑥3 𝑥𝑛 𝑓(𝑥) = 1 + 𝑥 (1) + 1 + 1 + … ….. + 1 + … …. 2! 3! 𝑛! 𝑥 𝑥2 𝑥3 𝑥4 𝑥𝑛 𝑓(𝑥) = 𝑒 = 1 + 𝑥 + + + ……..+ + … …. 2! 3! 4! 𝑛! ********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example : Expansion of e−x Solution : 𝐿𝑒𝑡 𝑓(𝑥) = 𝑒 −𝑥 by Maclaurin′ s Theorem: ′ (0) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 + 𝑓 (0) + 𝑓 (0) + ……..+ 𝑓 (0) + … …. 2! 3! 𝑛! 𝑓(𝑥) = 𝑒 −𝑥 ∴ 𝑓(0) = 𝑒 0 =1 𝑓 ′ (𝑥) = −𝑒 −𝑥 ∴ 𝑓 ′ (0) = −𝑒 0 = −1 𝑓 ′′ (𝑥) = 𝑒 −𝑥 ∴ 𝑓 ′′ (0) = 𝑒 0 =1 𝑓 ′′′ (𝑥) = −𝑒 −𝑥 ∴ 𝑓 ′′′ (0) = −𝑒 0 = −1 𝑓 ′𝑣 (𝑥) = 𝑒 −𝑥 ∴ 𝑓 ′𝑣 (0) = 𝑒 0 =1 𝑓 𝑣 (𝑥) = −𝑒 −𝑥 ∴ 𝑓 𝑣 (0) = −𝑒 0 = −1 ………….. 𝑓 𝑛 (𝑥) = (−1)𝑛 𝑒 −𝑥 ∴ 𝑓 𝑛 (0) = (−1)𝑛 𝑒 0 = (−1)𝑛 by Maclaurin′ s Theorem: ′ (0) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 + 𝑓 (0) + 𝑓 (0) + … ….. + 𝑓 (0) + … …. 2! 3! 𝑛! 𝑥2 𝑥3 𝑥𝑛 𝑓(𝑥) = 1 + 𝑥 (−1) + 1+ (−1) + … ….. + (−1)𝑛 + … …. 2! 3! 𝑛! −𝑥 𝑥2 𝑥3 𝑥4 𝑛 𝑥𝑛 𝑓(𝑥) = 𝑒 = 1 − 𝑥 + − + − … ….. +(−1) + … …. 2! 3! 4! 𝑛! ********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example : Expansion of sin x Solution : 𝐿𝑒𝑡 𝑓(𝑥) = sin 𝑥 by Maclaurin′ s Theorem: ′ (0) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 + 𝑓 (0) + 𝑓 (0) + ……..+ 𝑓 (0) + … …. 2! 3! 𝑛! 𝑓(𝑥) = sin 𝑥 ∴ 𝑓(0) = sin 0 =0 𝑓 ′ (𝑥) = cos 𝑥 ∴ 𝑓 ′ (0) = cos 0 = 1 𝑓 ′′ (𝑥) = − sin 𝑥 ∴ 𝑓 ′′ (0) = − sin 0 =0 𝑓 ′′′ (𝑥) = − cos 𝑥 ∴ 𝑓 ′′′ (0) = − cos 0 = −1 𝑓 ′𝑣 (𝑥) = sin 𝑥 ∴ 𝑓 ′𝑣 (0) = sin 𝑥 =0 𝑓 𝑣 (𝑥) = cos 𝑥 ∴ 𝑓 𝑣 (0) = cos 0 = 1 by Maclaurin′ s Theorem: 𝑥 2 ′′ 𝑥3 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 ′ (0) + 𝑓 (0) + 𝑓 ′′′ (0) + … ….. + 𝑓 (0) + … …. 2! 3! 𝑛! 𝑥2 𝑥3 𝑥4 𝑥5 𝑓(𝑥) = 0 + 𝑥 (1) + (0) + (−1) + (0) + (1) … …. 2! 3! 4! 5! 𝑥 3 𝑥 5 𝑥 7 𝑥 9 𝑥 11 𝑓(𝑥) = sin 𝑥 = 𝑥 − + − + − − … ….. 3! 5! 7! 9! 11! *********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Example : Expansion of sinh x Solution : 𝐿𝑒𝑡 𝑓(𝑥) = sinh x by Maclaurin′ s Theorem: ′ (0) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 + 𝑓 (0) + 𝑓 (0) + … ….. + 𝑓 (0) + … …. 2! 3! 𝑛! 𝑓(𝑥) = sinh x ∴ 𝑓(0) = sin 0 =0 𝑓 ′ (𝑥) = cosh 𝑥 ∴ 𝑓 ′ (0) = cos 0 = 1 𝑓 ′′ (𝑥) = sinh x ∴ 𝑓 ′′ (0) = sin 0 =0 𝑓 ′′′ (𝑥) = cosh 𝑥 ∴ 𝑓 ′′′ (0) = cos 0 = 1 𝑓 ′𝑣 (𝑥) = sinh x ∴ 𝑓 ′𝑣 (0) = sin 𝑥 =0 𝑓 𝑣 (𝑥) = cosh 𝑥 ∴ 𝑓 𝑣 (0) = cos 0 = 1 by Maclaurin′ s Theorem: ′ (0) 𝑥 2 ′′ 𝑥 3 ′′′ 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 + 𝑓 (0) + 𝑓 (0) + … ….. + 𝑓 (0) + … …. 2! 3! 𝑛! 𝑥2 𝑥3 𝑥4 𝑥5 𝑓(𝑥) = 0 + 𝑥 (1) + (0) + (−1) + (0) + (1) … …. 2! 3! 4! 5! 𝑥 3 𝑥 5 𝑥 7 𝑥 9 𝑥 11 𝑓(𝑥) = sin 𝑥 = 𝑥 + + + + + − … ….. 3! 5! 7! 9! 11! ********** 𝐒𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐄𝐱𝐩𝐚𝐧𝐬𝐢𝐨𝐧𝐬: x2 x3 x4 x5 1) ex = 1 + x + + + + + ………… 2! 3! 4! 5! x2 x3 x4 x5 2) e−x = 1 − x + − + − + ………… 2! 3! 4! 5! x3 x5 x7 x9 3) sin x = x − + − + − ………… 3! 5! 7! 9! x3 x5 x7 x9 4) sinh x = x + + + + + ………… 3! 5! 7! 9! x2 x4 x6 x8 5) cos x = 1 − + − + − ………… 2! 4! 6! 8! x2 x4 x6 x8 6) cosh x = 1 + + + + + ………… 2! 4! 6! 8! "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 x3 2x5 17x7 7) tan x = x + + + + ………… 3 15 315 x3 2x5 17x7 8) tanh x = x − + − + ………… 3 15 315 x2 x3 x4 x5 9) log(1 + x) = x − + − + − ………… 2 3 4 5 x2 x3 x4 x5 10) log(1 − x) = −x − − − − − ………… 2 3 4 5 n(n−1)x2 n(n−1)(n−2)x3 11) (1 + x)n = 1 + nx + + + ……………… 2! 3! 1 12) = (1 + x)−1 = 1 − x + x 2 − x 3 + x 4 − x 5 + … … … … … (1+x) 1 13) = (1 − x)−1 = 1 + x + x 2 + x 3 + x 4 + x 5 + … … … … … (1−x) 1 x3 1 3 x5 1 3 5 x7 14) sin−1 x = x + + + + ………… 2 3 24 5 246 7 1 x3 1 3 x5 1 3 5 x7 15) sinh−1 x = x − + − + ………… 2 3 24 5 246 7 π 1 x3 1 3 x5 1 3 5 x7 16) cos −1 x = – [x + + + + …………] 2 2 3 24 5 246 7 x3 x5 x7 17) tan−1 x = x − + − + ………… 3 5 7 x3 x5 x7 18) tanh−1 x = x + + + + ………… 3 5 7 ********* "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Examples: Expand of ex cos x in ascending powers of x upto a term in 𝑥 4 Solution: Let f(x) = ex cos x We know that x x2 x3 x4 e = 1 + x + + + + ………… 2! 3! 4! x2 x4 x6 x8 cos x = 1 − + − + − … … … … 2! 4! 6! 8! x x2 x3 x4 x2 x4 f(x) = e cos x = (1 + x + + + + ⋯ ) (1 − + … … … … ) 2! 3! 4! 2! 4! x2 x4 x2 x4 x2 x2 x4 ex cos x = 1 (1 − + … ) + x (1 − + …)+ (1 − 2! + 4! … ) + 2! 4! 2! 4! 2! x3 x2 x4 x4 x2 x4 3! (1 − 2! + 4! … ) + 4! (1 − 2! + 4! … ) x2 x4 x2 x4 x2 x2 x4 ex cos x = 1 (1 − + … ) + x (1 − + …)+ (1 − + …) + 2 24 2 24 2 2 24 x3 x2 x4 x4 x2 x4 6 (1 − 2 + 24 …)+ 24 (1 − 2 + 24 …) x x2 x4 x3 x2 x4 x3 x4 e cos x = 1 − + +x− + − + + 2 24 2 2 4 6 24 x3 x4 ex cos x = 1 + x − − 3 6 OR Examples: Expand of 𝑒 𝑥 cos x in ascending powers of x upto a term in 𝑥 4 Solution: 𝐿𝑒𝑡 𝑓(𝑥) = 𝑒 𝑥 cos x by Maclaurin′ s Theorem 𝑥 2 ′′ 𝑥3 𝑥 4 ′𝑣 𝑥𝑛 𝑛 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 ′ (0) + 𝑓 (0) + 𝑓 ′′′ (0) + 𝑓 (0) … ….. + 𝑓 (0) + … …. 2! 3! 4! 𝑛! 𝑓(𝑥) = 𝑒 𝑥 cos x ∴ 𝑓(0) = 𝑒 0 cos 0 =1 𝑓 ′ (𝑥) = 𝑒 𝑥 cos x − 𝑒 𝑥 sin x ∴ 𝑓 ′ (0) = 𝑒 0 cos 0 − 𝑒 0 sin 0 = 1 𝑓 ′′ (𝑥) = 𝑒 𝑥 cos x − 𝑒 𝑥 sin x − 𝑒 𝑥 cos x − 𝑒 𝑥 sin x = −2𝑒 𝑥 sin x ∴ 𝑓 ′′ (0) = 0 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝑓 ′′′ (𝑥) = −2𝑒 𝑥 sin x − 2𝑒 𝑥 cos x ∴ 𝑓 ′′′ (0) = −2 𝑓 ′𝑣 (𝑥) = −2𝑒 𝑥 sin x − 2𝑒 𝑥 cos x + 2𝑒 𝑥 sin x − 2𝑒 𝑥 cos x ∴ 𝑓 ′𝑣 (0) = −4 by Maclaurin′ s Theorem: 𝑥2 𝑥3 𝑥4 𝑓(𝑥) = 𝑓(0) + 𝑥 𝑓 ′ (0) + 𝑓 ′′ (0) + 𝑓 ′′′ (0) + 𝑓 ′𝑣 (0) + … …. 2! 3! 4! 𝑥2 𝑥3 𝑥4 𝑓(𝑥) = 1 + 𝑥 (1) + (0) + (−2) + (−4) 2 6 24 𝑥3 𝑥4 𝑓(𝑥) = 1 + 𝑥 − − 3 6 *********** Examples : Expand √1 + sin 𝑥 in ascending powers of x upto a term in 𝑥 6 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧 ∶ 𝐿𝑒𝑡 𝑓(𝑥) = √1 + sin 𝑥 𝑥 𝑥 𝑥 𝑥 𝑓(𝑥) = √sin2 ( ) + cos 2 ( ) + 2 sin ( ) cos ( ) 2 2 2 2 𝑥 𝑥 2 𝑓(𝑥) = √(sin ( ) + cos ( )) ∵ (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 2 2 𝑥 𝑥 𝑓(𝑥) = √1 + sin 𝑥 = sin ( ) + cos ( ) 2 2 We know that x3 x5 x7 x9 sin x = x − + − + − ………… 3! 5! 7! 9! x2 x4 x6 x8 cos x = 1 − + − + − ………… 2! 4! 6! 8! x Put x = in above expansion 2 x x 1 x 3 1 x 5 sin ( ) = − ( ) + ( ) − … … … … 2 2 3! 2 5! 2 x 1 x 2 1 x 4 cos ( ) = 1 − ( ) + ( ) − … … … … 2 2! 2 4! 2 𝑥 𝑥 𝑓(𝑥) = √1 + sin 𝑥 = sin ( ) + cos ( ) 2 2 x 1 x 3 1 x 5 1 x 2 1 x 4 1 x 6 𝑓(𝑥) = − ( ) + ( ) − … … … + 1 − ( ) + ( ) − ( ) … … 2 3! 2 5! 2 2! 2 4! 2 6! 2 3 5 2 4 x 1 x 1 x 1 x 1 x 1 x6 𝑓(𝑥) = − ∗ + ∗ − ……… + 1 − ∗ + ∗ − ∗ …… 2 6 8 120 32 2 4 24 16 720 64 x x2 x3 x4 x5 x6 𝑓(𝑥) = 1 + − − + + − 2 8 48 384 3840 46080 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Type: Expansions of functions by using substitution:- 2𝑥 Ex.1 – Expand sin−1 ( ) in ascending powers of x 1 + 𝑥2 Solution: 2x Let f(x) = sin−1 ( ) 1 + x2 Put x = tanθ (∴ θ = tan−1 x) 2 tan θ ∴ f(x) = sin−1 ( ) 1 + tan2 θ = sin−1 (sin 2θ) =2θ =2(tan−1 x) (If x = tanθ then θ = tan−1 x) x3 x5 x7 We know that tan−1 x = x − + − + ………… 3 5 7 2x x3 x5 x7 ∴ sin−1 ( ) =2 [ x − + − + … … … …] 1+x2 3 5 7 ********* 1 1 x3 1 3 x5 Ex.2- Prove that sec −1 [ 2 ] = 2 [x + + …..] 1−2𝑥 2 3 24 5 Solution: 1 Let 𝑓(𝑥) = sec −1 [ ] 1 − 2𝑥 2 Put 𝑥 = sin 𝜃 (∴ 𝜃 = sin−1 𝑥) 1 ∴ 𝑓(𝑥) = sec −1 [ ] 1 − 2𝑠𝑖𝑛2 𝜃 1 =sec −1 [ ] cos 2𝜃 = sec −1 [sec 2𝜃] =2𝜃 (𝐼𝑓 𝑥 = 𝑠𝑖𝑛𝜃 𝑡ℎ𝑒𝑛 𝜃 = sin−1 𝑥) =2 sin−1 𝑥 1 1 x3 1 3 x5 sec −1 [ 2 ] = 2 [x + + ….. ] Hence Proved. 1−2𝑥 2 3 24 5 ********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 Ex.3 – Expand cos −1 (4𝑥 3 − 3𝑥) in ascending powers of x. Solution: Let 𝑓(𝑥) = cos −1 (4𝑥 3 − 3𝑥) Put 𝑥 = 𝑐𝑜𝑠𝜃 𝑓(𝑥) = cos −1 (4𝑐𝑜𝑠 3 𝜃 − 3𝑐𝑜𝑠𝜃) = cos −1 (𝑐𝑜𝑠3𝜃) =3𝜃 = 3 cos −1 𝑥 (𝐼𝑓 𝑥 = 𝑐𝑜𝑠𝜃 𝑡ℎ𝑒𝑛 𝜃 = cos −1 𝑥 ) π 1 x3 1 3 x5 1 3 5 x7 = 3 [ – (x + + + + … … … … )] 2 2 3 24 5 246 7 π 1 x3 1 3 x5 1 3 5 x7 =3 – 3 [x + + + + …………] 2 2 3 24 5 246 7 ********** 3𝑥 − 𝑥 3 𝜋 x3 x5 x7 Ex.4- Prove that cot −1 ( ) = 2 − 3 (x − + − + …………) 1 − 3𝑥 2 3 5 7 x3 3x5 Ex.5- Prove that sin−1 (3𝑥 − 4𝑥 3 ) = 3 (x + + + …………) 6 40 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐈𝐧𝐝𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐚𝐭𝐞 𝐅𝐨𝐫𝐦𝐬 𝐃𝐞𝐟𝐢𝐧𝐚𝐭𝐢𝐨𝐧: Let f(x) and g(x) be any two function of x such that f(a) = 0 and g(a) = 0 f(x) 0 then the ratio is said to be the indeterminate form at x = a g(x) 0 0 ∞ There are several Indeterminate Forms , , 0 × ∞, ∞ − ∞, 00 , ∞0 , 1∞ 0 ∞ 𝐓𝐫𝐮𝐞 𝐕𝐚𝐥𝐮𝐞 (𝐋𝐢𝐦𝐢𝐭): The limiting value of an indeterminate form is called its true value. 𝟎 𝐓𝐲𝐩𝐞 𝐈 ∶ 𝐈𝐧𝐝𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐚𝐭𝐞 𝐅𝐨𝐫𝐦 (𝐋′ 𝐇𝐨𝐬𝐩𝐢𝐭𝐚𝐥 𝐑𝐮𝐥𝐞) 𝟎 Let f(x) and g(x) be any two function of x such that f(a) = 0 and g(a) = 0 f(x) f ′ (x) If lim f(x) = 0 and lim g(x) = 0 then lim = lim ′ x→a x→a x → a g(x) x → a g (x) ∞ 𝐓𝐲𝐩𝐞 𝐈𝐈 ∶ 𝐈𝐧𝐝𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐚𝐭𝐞 𝐅𝐨𝐫𝐦 ∞ f(x) ∞ 0 If lim f(x) = ∞ and lim g(x) = ∞ then lim in form then reduces to by x→a x→a x→a g(x) ∞ 0 f(x) 1/f(x) = and L′ Hospital Rule is applicable. g(x) 1/g(x) ′ ∞ f(x) f ′ (x) L Hospital Rule is applied to the form Thus lim = lim ′ ∞ x → a g(x) x → a g (x) 𝐓𝐲𝐩𝐞 𝐈𝐈𝐈 ∶ 𝐈𝐧𝐝𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐚𝐭𝐞 𝐅𝐨𝐫𝐦 𝟎 × ∞ If lim f(x) = 0 and lim g(x) = ∞ then lim f(x) ∙ g(x) takes 0 × ∞ x→a x→a x→a f(x) g(x) 0 ∞ f(x) ∙ g(x) = 1 or 1 and the limit reduces to either form or form 0 ∞ g(x) f(x) and L′ Hospital Rule is applicable. 𝐓𝐲𝐩𝐞 𝐈𝐕 ∶ 𝐈𝐧𝐝𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐚𝐭𝐞 𝐅𝐨𝐫𝐦 ∞ − ∞ If lim f(x) = ∞ and lim g(x) = ∞ then lim [f(x) − g(x)] takes ∞ − ∞ x→a x→a x→a 0 ∞ simplify the expression f(x) − g(x) and the limit reduces to either form or form 0 ∞ and L′ Hospital Rule is applicable. "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐍𝐨𝐭𝐞: 𝐟𝟏 (𝐱) 𝐠 𝟏 (𝐱) 𝐟𝟏 (𝐱)𝐠 𝟐 (𝐱) − 𝐠 𝟏 (𝐱)𝐟𝟐 (𝐱) 0 1) If 𝐥𝐢𝐦 − in ∞ − ∞ form then 𝐥𝐢𝐦 is in 𝐱 → 𝐚 𝐟𝟐 (𝐱) 𝐠 𝟐 (𝐱) 𝐱→𝐚 𝐟𝟐 (𝐱)𝐠 𝟐 (𝐱) 0 2) If f ′ (x), f ′′ (x) … … f n−1 (x) and g ′ (x), g ′′ (x) … … g n−1 (x) all are zero, f(x) fn (x) But f n (x)and g n (x)are not both zero then lim = lim x → a g(x) x → a gn (x) 3) Use of L′ Hospital Rule ∶ Differentiate numerator and denominator separately and then put x = a. If this reduces to indeterminate form then apply the rule again. 4) If logrithmic term is present in 0 ∗ ∞ form then keep logritmic term in numerator 𝐓𝐲𝐩𝐞 𝐕 ∶ 𝐈𝐧𝐝𝐞𝐭𝐞𝐫𝐦𝐢𝐧𝐚𝐭𝐞 𝐅𝐨𝐫𝐦 𝟎𝟎 , ∞𝟎 , 𝟏∞ 1) If lim f(x) = 0 and lim g(x) = 0 then lim {f(x)}g(x) takes 00 x→a x→a x→a 2) If lim f(x) = ∞ and lim g(x) = 0 then lim {f(x)}g(x) takes ∞0 x→a x→a x→a 3) If lim f(x) = 1 and lim g(x) = ∞ then lim {f(x)}g(x) takes 10 x→a x→a x→a If the true value of limit is denoted by L Then L = lim {f(x)}g(x) x→a Taking Log on both sides log L = log lim {f(x)}g(x) x→a log L = lim g(x) log f(x) x→a limit can be determined by 0 × ∞ form, true value is b log L = b L = eb 𝐍𝐨𝐭𝐞: e∞ = ∞ e−∞ = 0 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐒𝐭𝐚𝐧𝐝𝐚𝐫𝐝 𝐋𝐢𝐦𝐢𝐭𝐬: sin x tan x 1) lim =1 3) lim =1 6) lim (1 + x)1/x = e x→0 x x→0 x x→0 sin−1 x tan−1 x 1 x 2) lim =1 4) lim =1 7) lim (1 + ) = e x→0 x x→0 x x →∞ x sinh x ex − 1 x a −1 3) lim =1 5) lim =1 8) lim = log a x→0 x x→0 x x→0 x ********** 𝐱𝐞𝐱 − 𝐥𝐨𝐠(𝟏 + 𝐱) 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟏) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 𝐱→𝟎 𝐱𝟐 xex − log(1 + x) 𝐒𝐨𝐥𝐮𝐭𝐨𝐢𝐧: Let L = lim … … … … (1) x→0 x2 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 (0)e0 − log(1 + 0) 0(1) − log 1 0−0 0 0 L = = = = ∴ form 02 0 0 0 0 1 xex + ex − (1 + x) From (1) L = lim … … … … (2) x→0 2 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 1 1 (0)e0 + e0 − 0(1) + 1 − L = 1+0 = 1−0 = 0 ∴ 0 form 2(0) 2(0) 0 0 −1 xex + ex + ex − (1 + x)2 From (2) L = lim x→0 2 1 xex + ex + ex + (1 + x)2 L = lim … … … … (3) x→0 2 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 1 (0)e0 + e0 + e0 + (1 + 0)2 L = 2 0+1+1+1 3 L= = 2 2 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐜𝐨𝐬 𝟐 (𝛑𝐱) 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟐) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 𝟐𝐱 𝐱 → 𝐞 − 𝟐𝐱𝐞 𝟏 𝟐 cos 2 πx 𝐒𝐨𝐥𝐮𝐭𝐨𝐢𝐧: Let L = lim 2x … … … … (1) x → e − 2xe 1 2 𝟏 𝟏 𝐱 → 𝐩𝐮𝐭 𝐱 = 𝟐 𝟐 cos 2 π(1/2) 0 0 L = 2(1/2) = = e − 2(1/2)e e−e 0 2 cos πx (− sin πx) π From (1) L = lim ∵ sin 2x = 2 sin x cos x x→ 1 2e2x − 2e 2 − π sin(2πx) L = lim … … … … (2) x→ 1 2e2x − 2e 2 𝟏 𝟏 𝐱 → 𝐩𝐮𝐭 𝐱 = 𝟐 𝟐 − π sin(2π/2) − π(0 ) 0 L = = = 2e2(1/2) − 2e 2e − 2e 0 – π cos(2πx) (2π) L = lim 2x − 0 … … … … (3) x→ 1 4e 2 𝟏 𝟏 𝐱 → 𝐩𝐮𝐭 𝐱 = 𝟐 𝟐 – 2π2 cos(2π/2) L = 4e2(1/2) – 2π2 (−1) L = 4e 2π2 π2 L = = 4e 2e ********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐥𝐨𝐠 𝐬𝐢𝐧 𝟐𝐱 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟑) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 𝐱→𝟎 𝐥𝐨𝐠 𝐬𝐢𝐧 𝐱 log sin 2x 𝐒𝐨𝐥𝐮𝐭𝐨𝐢𝐧: Let L = lim … … … … (1) x→0 log sin x 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 log sin 2(0) log 0 ∞ L = = = from log sin (0) log 0 ∞ 1 cos 2x (2) d 1 From (1) L = lim sin 2x ∵ log x = x→0 1 dx x cos x sin x 2 cot 2x cos x 1 L = lim ∵ = cot x = tan 𝑥 x → 0 cot x sin x cot x 2 2 tan x L = lim tan 2x = lim … … … (2) x→0 1 x → 0 tan 2x tan x 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 2 tan 0 0 L = = from tan 20 0 2 sec 2 x 2 sec 2 0 1 L = lim = = =1 x → 0 sec 2 2x (2) 2 sec 2 2(0) 1 OR 𝐥𝐨𝐠 𝐬𝐢𝐧 𝟐𝐱 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟑) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 𝐱→𝟎 𝐥𝐨𝐠 𝐬𝐢𝐧 𝐱 log sin 2x 𝐒𝐨𝐥𝐮𝐭𝐨𝐢𝐧: Let L = lim … … … … (1) x→0 log sin x 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 log sin 2(0) log 0 ∞ L = = form log sin (0) log 0 ∞ 1 cos 2x (2) L = lim sin 2x x→0 1 cos x sin x "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 2 cos 2x sin x L = lim x → 0 sin 2x cos x 2 cos 2x sin x L = lim ∵ sin 2x = 2 sin x cos x x → 0 2 sin x cos x cos x cos 2x L = lim ∵ cos 2x = cos 2 x − sin2 x x → 0 cos 2 x cos 2 x − sin2 x L = lim x→0 cos 2 x cos 2 x sin2 x L = lim − x → 0 cos 2 x cos 2 x L = lim 1 − tan2 𝑥 x→0 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 L = 1 − tan2 0 L=1−0=1 ∗∗∗∗∗∗∗∗∗∗∗ 𝐞𝐱 − 𝟏 − 𝐱 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟒) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 𝐱 → 𝟎 𝐥𝐨𝐠 (𝟏 + 𝐱) − 𝐱 ex − 1 − x 𝐒𝐨𝐥𝐮𝐭𝐨𝐢𝐧: Let L = lim … … … … (1) x → 0 log (1 + x) − x 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 e0 − 1 − 0 1−1 0 L = = form log (1 + 0) − 0 log (1) 0 ex − 0 − 1 From (1) L = lim … … … … (2) x→0 1 −1 1+x e0 − 0 − 1 0 L= form 1 0 −1 1+0 ex From (2) L = lim … … … … (3) x→0 −1 (1 + x)2 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 e0 L= −1 = −1 (1+0)2 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐱 (𝟏 + 𝐚 𝐜𝐨𝐬 𝐱) − 𝐛 𝐬𝐢𝐧 𝐱 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟓) 𝐅𝐢𝐧𝐝 𝐚 𝐚𝐧𝐝 𝐛 𝐢𝐟 𝐥𝐢𝐦 =𝟏 𝐱→𝟎 𝐱𝟑 x(1 + acos x) − bsin x 𝐒𝐨𝐥𝐮𝐭𝐨𝐢𝐧: Let L = lim … … … … (1) x→0 x3 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 (0)(1 + a cos 0) − b sin 0 0 L = 3 = form 0 0 x (0 − a sin x) + (1 + a cos x) − bcos x From (1) L = lim x→0 3x 2 −a x sin x + 1 + a cos x − bcos x L = lim … … … … (2) x→0 3x 2 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 0 (0 − a sin 0) + (1 + a cos 0) − b cos 0 (1 + a ) − b L = = =∞ 3(0)2 0 (1 + a ) − b but limit is finite i. e. 1 1≠ =∞ 0 0 (1 + a ) − b 0 ∴ For Finite limit it should be from = from 0 0 0 ∴ (1 + a ) − b = 0 ∴ a − b = −1 … … … (A) – a x cos x − a sin x + 0 − a sin x + b sin x From (2) L = lim … … … … (3) x→0 6x 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 – a (0) cos (0) − a sin(0) − a sin (0) + b sin(0) 0 L = = 6(0) 0 a x sin x – a cos x − a cos x − a cos x + b cos x From (3) L = lim … … … … (4) x→0 6 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 a(0) sin(0) – a cos (0) − a cos(0) − a cos (0) + b cos(0) L = 6 0 –a− a− a + b L = 6 – 3a + b L = but limit value L = 1 is given 6 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 – 3a + b but limit is finite 1 ∴1 = 6 ∴ – 3a + b = 6 … … … (B) ∴ a − b = −1 … … … (A) Solving (A) and (B) a − b = −1 – 3a + b = 6 5 Add (A) and (B) −2a=5 ∴ a=− 2 5 5 3 Put a = − in (A) ∴ − −b=1 ∴b=− 2 2 2 5 3 ∴a=− 𝑏=− 2 2 ********** 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟔) 𝐅𝐢𝐧𝐝 𝐚 𝐚𝐧𝐝 𝐛 𝐢𝐟 𝐥𝐢𝐦 [𝐱 −𝟑 𝐬𝐢𝐧 𝐱 + 𝐚𝐱 −𝟐 + 𝐛] = 𝟎 𝐱→𝟎 𝐒𝐨𝐥𝐮𝐭𝐨𝐢𝐧: Let L = lim [x −3 sin x + ax −2 + b] x→0 sin x a L = lim [ 3 + 2 + b] x→0 𝑥 𝑥 sin x + ax + bx 3 L = lim [ ] … … …. (1) x→0 𝑥3 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 sin (0) + a(0) + b(0)3 0 L= = form (0)3 0 cos x + a + 3x 2 b From (1) L = lim [ ] … … …. (2) x→0 3𝑥 2 cos (0) + a + 3x(0)2 b 1+a L= = =∞ 3(0)2 0 but limit is finite L = 0 0 1+a 0 ∴ For Finite limit it should be ∴ = 0 0 0 ∴ 1+a=0 ∴ a = −1 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 −sin x + 6xb From (2) L = lim [ ] … … …. (3) x→0 6𝑥 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 −sin (0) + 6(0)b 0 L= form 6(0) 0 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 −cos x + 6b From (3) L = lim [ ] … … …. (4) x→0 6 −cos (0) + 6b 𝐿= 6 −1 + 6b 𝐿= 6 −1 + 6b but limit is finite 0 ∴ =0 6 1 ∴ −1 + 6b = 0 𝑏= 6 1 ∴ a = −1 𝑏= 6 ********** a cos x − a + bx 2 1 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟕) Find a and b if lim = x→0 x4 12 a cos x − a + bx 2 𝐒𝐨𝐥𝐮𝐭𝐨𝐢𝐧: Let L = lim … … …. (1) x→0 x4 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 a cos (0) − a + b(0)2 a −a 0 L= = form (0)4 0 0 −a sin x + 2xb From (1) L = lim … … …. (2) x→0 4x 3 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 −a sin (0) + 2(0)b 0 L= form 4(0)3 0 – a cos x + 2b From (2) L = lim [ ] … … …. (3) x→0 12𝑥 2 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 – a cos(0) + 2b −a + 2b 1 L= = ≠ 12(0)2 0 12 1 but limit is finite 12 0 ∴ For Finite limit it should be 0 ∴ −a + 2b = 0 … … …. (A) a sin x From (3) L = lim [ ] … … …. (3) x→0 24𝑥 a sin x L= lim [ ] 24 x → 0 𝑥 1 a = (1) 12 24 𝑎=2 ∴ −a + 2b = 0 ∴ −2 + 2b = 0 𝑏=1 𝑎 = 2 and 𝑏 = 1 ******** a sin2 x + b log cos x 1 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟖) Find a and b if lim = − x→0 x4 2 sin 2x + p sin x 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟗) If lim is finite then find the value of p x→0 x3 and hence find the value of limit ************** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟏𝟎) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 𝐬𝐢𝐧 𝐱 𝐥𝐨𝐠 𝐱 𝐱→𝟎 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Let L = lim sin x log x … … … … (1) x→0 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 L = sin 0 log 0 𝟎 × ∞ 𝐟𝐨𝐫𝐦 log x log x L = lim = lim … … … …. (2) x→0 1 x → 0 cosec x sin 𝑥 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 log 0 ∞ L= 𝐟𝐨𝐫𝐦 cosec 0 ∞ 1/x From (2) L = lim x → 0 −cosec x cotx sinx tan x L = lim x→0 −x sinx tan x 𝟎 L = lim 𝐟𝐨𝐫𝐦 … … … … …. (𝟑) x→0 −x 𝟎 sin x sec 2 x + tan x cos x From (3) L = lim … … … … … (4) x→0 −1 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 sin(0) sec 2 (0) + tan(0) cos(0) L= −1 L= 0 ********** "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟏𝟏) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 (𝟏 − 𝐬𝐢𝐧 𝐱)𝐭𝐚𝐧 𝐱 𝐱 → 𝛑/𝟐 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Let L = lim (1 − sin x) tan x … … … … (1) x → π/2 𝐱 → 𝛑/𝟐 𝐩𝐮𝐭 𝐱 = 𝛑/𝟐 L = (1 − sin π/2)tan π/2 0 × ∞ form (1 − sin x) (1 − sin x) From (1) L = lim = lim … … ….. (2) 𝐱 → 𝛑/𝟐 1 𝐱 → 𝛑/𝟐 cot x tan x 𝐱 → 𝛑/𝟐 𝐩𝐮𝐭 𝐱 = 𝛑/𝟐 (1 − sin π/2) 0 L= form cot π/2 0 – cos x From (2) L = lim 𝐱 → 𝛑/𝟐 −cosec 2 x 𝐱 → 𝛑/𝟐 𝐩𝐮𝐭 𝐱 = 𝛑/𝟐 – cos π/2 L= −cosec 2 π/2 0 L= 12 L= 0 ***************** 𝛑 𝛑 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟏𝟐) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 [ − ] 𝐱 → 𝟎 𝟒𝐱 𝟐𝐱(𝐞𝛑𝐱 + 𝟏) π π 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧 ∶ Let L = lim [ − ] … … … … (1) x → 0 4x 2x(eπx + 1) 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 π π L= − ∞ − ∞ form 4(0) 2(0)(eπ0 + 1) 2xπ(eπx + 1) − 4xπ From (1) L = lim [ ] x→0 8x 2 (eπx + 1) π(eπx + 1) − 2π L = lim [ ] x→0 4x(eπx + 1) "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 πeπx + π − 2π L = lim [ ] x→0 4x(eπx + 1) πeπx − π L = lim [ ] … … ….. (2) x→0 4x(eπx + 1) 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 πe0x − π π −π 𝟎 L= = 𝐟𝐨𝐫𝐦 4(0)(eπ0 + 1) 0 𝟎 π2 eπx From (2) L = lim [ ] x→0 4x(πeπx + 0) + 4(πeπx + 1) π2 eπ0 L= 4(0)(πe0 + 0) + 4(e0 + 1) π2 L= 4(1 + 1) π2 L= 8 ******************* 𝟏 𝟏 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟏𝟑) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 [ − 𝟐 𝐥𝐨𝐠 (𝟏 + 𝐱)] 𝐱→𝟎 𝐱 𝐱 Solution: 1 1 Let L = lim [ − 2 log (1 + x)] … … … … (1) x→0 x x 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 1 1 0 L= − log(1 + 0) ∞ − form 0 0 0 x 1 From (1) L = lim [ 2 − 2 log (1 + x) ] x→0 x x x ∗ x 2 − x 2 log(1 + x) L = lim [ ] x→0 x2x2 x 2 (x − log(1 + x)) L = lim [ ] x→0 x2x2 x − log(1 + x) 0 L = lim [ ] form … … … … (1) x→0 x2 0 "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 1 1− From (1) L = lim [ 1+x] … … … (2) x→0 2x 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 1 1− 𝟎 L= 1+0 𝐟𝐨𝐫𝐦 2(0) 𝟎 −1 0− (1 + x)2 From (2) L = lim [ ] x→0 2 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 1 (1 + 0)2 L = 2 1 L= 2 ******************** 𝟏 𝟐 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟏𝟒) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 [ − 𝟐 ] 𝐱→𝟏 𝐱 − 𝟏 𝐱 −𝟏 1 2 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧 ∶ Let L = lim [ − 2 ] … … … … (1) x→1 x − 1 x −1 𝐱 → 𝟏 𝐩𝐮𝐭 𝐱 = 𝟏 1 2 1 2 L= − 2 = − ∞ − ∞ 𝐟𝐨𝐫𝐦 1−1 1 −1 0 0 (x 2 − 1) − 2(x − 1) From (1) L = lim [ ] x→0 (x − 1)(x 2 − 1) (x − 1)(x + 1) − 2(x − 1) L = lim [ ] x→0 (x − 1)(x 2 − 1) (x + 1) − 2 L = lim [ ] x→0 (x 2 − 1) x−1 L = lim [ 2 ] x→0 (x − 1) "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 x−1 L = lim [ ] x→0 (x − 1)(x + 1) 1 L = lim [ ] x→0 (x + 1) 𝐱 → 𝟏 𝐩𝐮𝐭 𝐱 = 𝟏 1 L= 1+1 1 L= 2 *********** 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟏𝟓) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 {𝐬𝐢𝐧 𝐱}𝐭𝐚𝐧 𝐱 𝐱→𝟎 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Let L= lim {sin x}tan x … … … … (1) x→0 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 L = {sin 0}tan 0 𝟎𝟎 𝐟𝐨𝐫𝐦 Taking log on both sides log L = log lim {sin x}tan x x→0 log L = lim log{sin x}tan x ∵ log an = n ∗ log a x→0 log L = lim tan x log sin x … … … … (2) x→0 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 log L = tan 0 log sin 0 = tan 0 log 0 𝟎 × ∞ 𝐟𝐨𝐫𝐦 log sin x From (2) log L = lim x→0 1/ tan x log sin x 1 log L = lim ∴ = cot 𝑥 … … … … (3) x→0 cot x tan x 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 log sin 0 ∞ log L = 𝐟𝐨𝐫𝐦 cot 0 ∞ 1 d cos x ( ) sin x From (3) log L = lim sin x dx2 = sin x 2 x→0 −cosec x −cosce x "𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢" Other Subjects: https://www.studymedia.in/fe/notes 𝐄𝐧𝐠𝐢𝐧𝐞𝐞𝐫𝐢𝐧𝐠 𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬 𝐈 cot x ∞ log L = lim 𝐟𝐨𝐫𝐦 … … …. (𝟒) x→0 −cosec 2 x ∞ −cosec 2 x d From (4) log L = lim ∵ x n = nx n−1 (x) x → 0 −2cosec 1 x cosec x cot x dx 1 1 log L = lim = lim tan x x → 0 2 cot x x→0 2 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 1 log L = tan 0 2 log L = 0 elog L = e0 L = e0 = 1 ****************** 𝐄𝐱𝐚𝐦𝐩𝐥𝐞 𝟏𝟔) 𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐞 𝐥𝐢𝐦 {𝐜𝐨𝐭 𝐱}𝐬𝐢𝐧 𝐱 𝐱→𝟎 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Let L= lim {cot x}sin x x→0 𝐱 → 𝟎 𝐩𝐮𝐭 𝐱 = 𝟎 L= lim {cot 0}sin 0 ∞𝟎 𝐟𝐨?