Limit-Süreklilik Konu Anlatımı PDF

## Limit :A→B is a function. lim f(x) = L means that when x is close to c, f(x) is close to L. For this to be true, c must be in the domain of f. However, c is not the point where we analyze the limit; the limit could be reached by x getting close to c without ever actually reaching that point. ###...

## Limit :A→B is a function. lim f(x) = L means that when x is close to c, f(x) is close to L. For this to be true, c must be in the domain of f. However, c is not the point where we analyze the limit; the limit could be reached by x getting close to c without ever actually reaching that point. ### Important Note: - The function must be defined in neighborhoods of c. - Defined for all points around c, but not necessarily at point c. ### Hyori: 1. From the definition, a function *must* be defined in the neighborhood of a point to have a limit at that point. However, it does not need to be defined *at* that point. 2. The value of the function at the point does not have to be equal to the limit. ### Definition: - The set (a-ε, a+ε) is called the ε-neighborhood of a. ### Definition: - If for every ε>0, there is at least one point in (a-ε, a+ε) which is in the set A but different from a, then a is called a *limit point* or *accumulation point*. ### Definition: - If A is a set, a is a point in A, and f:A→R is a function, then: - If for every ε>0, there exists a δ(ε)>0 such that 0<|x-a|<δ(ε) implies |f(x) - L| <ε for all x in A, then we say that L is the limit of f at a and we write: - lim f(x) = L - a→x ### Example: - Prove that lim (2x+5) = 7 - x→1 - **Solution:** Let ε>0. If |x-1|<ε/2, then |2x+5-7| = |2x-2| = 2|x-1| <2.(ε/2) = ε. - Therefore, lim (2x+5)=7. - x→1 - Prove that lim (x^2 + 3x) = 4 - x→1 - **Solution:** Let ε>0. If |x-1|<δ, then |x^2 + 3x - 4|=|x-1|*|x+4| <δ.(δ+5) <ε. - This inequality will be true when δ < (ε - 5 + √25 + 4ε)/2. - Therefore, lim (x^2 + 3x) = 4. - x→1 - Prove that lim x^2 = 0 - x→0 - **Solution:** Let ε>0. If |x-0|<ε, then |x^2 - 0| = |x^2| <ε. - Therefore, lim x^2 = 0 - x→0 ### Exercise: - Prove that lim (3x+1)=7 - x→2 ### Theorem - Let A⊂ℝ, f:A→ℝ, and p:A→ℝ be functions and a∈A. If lim f(x) = L and lim p(x) = M, then: - lim (cf)(x) = c*lim f(x) - x→a - x→a - lim (f+p)(x) = lim f(x) + lim p(x) - x→a - x→a - xa - lim (f*p)(x) = lim f(x) * lim p(x) - x→a - x→a - x→a - If for every x∈A, p(x)≠0 and lim p(x)≠0, then - x→a - lim f(x)/p(x) = lim f(x)/lim p(x) - x→a - x→a - x→a ### Example: - Calculate the limit: lim(5x^3 - 4x^2 + 2x - 3) - x→2 - **Solution:** lim(5x^3 - 4x^2 + 2x - 3) = lim 5x^3 - lim 4x^2 + lim 2x - lim 3 - x→2 - x→2 - x→2 - x→2 - = 5*(lim x^3) - 4*(lim x^2) + 2*(lim x) - lim 3 - x→2 - x→2 - x→2 - x→2 - = 5*(2^3) - 4*(2^2) + 2*(2) - 3 - = 40 - 16 + 4 - 3 - = 25 ### Definition: - If f is a function defined on (c,a), then: - If for every ε>0, there exists a δ>0 such that, c-δ<x<c implies |f(x)-L|<ε, then we say that L is the left-hand limit of f at c and we write: - lim f(x) = L - x→c- - If for every ε>0, there exists a δ>0 such that, c<x<c+δ implies |f(x)-L|<ε, then we say that L is the right-hand limit of f at c and we write: - lim f(x) = L - x→c+ ### Definition: - If lim f(x) = lim f(x) = L then lim f(x) = L - x→c- - x→c+ - x→c- ### Hyori: - If the function is only defined on the interval (a,b), and it approaches x from the left, then finding the limit is only possible from the left. ### Definition: - For every ε>0, there exists an N, such that for every x<N, |f(x)-L|<ε. ### Example: - Prove that lim 1/x = 0 - x→-∞ - **Solution:** Let ε>0. If x < -1/ε, then |1/x-0| = |-1/x| = 1/|x| <ε. - Therefore, lim 1/x = 0 - x→-∞ ### Exercise: - Calculate the limit: lim x/[x] - x→-∞ ### Example: - Calculate the limit: lim(5x^3 - 4x^2 + 3x + 2)/(2x^3 + x^2 + 5) - x→-∞ - **Solution:** lim(5x^3 - 4x^2 + 3x + 2)/(2x^3 + x^2 + 5) = lim(5 - 4/x + 3/x^2 + 2/x^3)/(2 + 1/x + 5/x^3) - x→-∞ - x→-∞ - = (5 - 0 + 0 + 0)/(2 + 0 + 0) - = 5/2 ### Example: - Calculate the limit: lim(3x+4)/(4x^2 + 5x + 2) - x→-∞ - **Solution:** lim(3x+4)/(4x^2 + 5x + 2) = lim (3/x + 4/x^2)/(4 + 5/x + 2/x^2) - x→-∞ - x→-∞ - = (0 + 0)/(4 + 0 + 0) - = 0 ### Example: - Calculate the limit: lim(x+3)/(2x^2+5) - x→-∞ - **Solution:** lim(x+3)/(2x^2+5) = lim (1/x + 3/x^2)/(2 + 5/x^2) - x→-∞ - x→-∞ - = (0 + 0)/(2 + 0) - = 0 ### Example: - Calculate the limit: lim √(x^2 + 2x) - x→∞ - Calculate the limit: lim √(x^2 + 2x) - x - x→∞ ### Definition: - If for every ε>0, there exists a δ>0, such that 0<|x-a|<δ implies |f(x)|>B, then we say that the limit of f(x) as x approach a, is positive infinity and we write: - lim f(x) = +∞ - x→a ### Definition:: - If for every K, there exists a δ>0, such that 0<|x-a|<δ implies f(x)<K, then we say that the limit of f(x) as x approach a, is negative infinity and we write: - lim f(x) = -∞ - x→a ### Theorem (Sandwich Theorem) - If p(x) ≤ f(x) ≤ h(x) and lim p(x) = lim h(x) = L, then lim f(x) = L. - x→a - x→a - x→a ### Example: - Calculate the limit: lim (x^2 + 1)*sin(1/x) - x→0 - **Solution:** We know that -1 ≤ sin(1/x) ≤ 1. - Therefore, -x^2 - 1 ≤ (x^2+1)*sin(1/x) ≤ x^2 + 1. - lim(-x^2 - 1) = lim(x^2 + 1) = 0. - x→0 - x→0 - So, using the Sandwich Theorem, we have lim (x^2 + 1)*sin(1/x) = 0. - x→0 ## Basic Trigonometric Limits - Consider the unit circle of radius 1. Let O be the angle θ. Let A(1,0), B and C be points on the circle as it is shown in the attached image. - Area(BOH) < Area(BOA) < Area(COA), so: - 1/2 * cos(θ) * sin(θ) < 1/2 * θ < 1/2 tan(θ). - Multiplying both sides by 2/sin(θ) we get: - cos(θ) < θ/sin(θ) < 1/cos(θ). - Taking the limits, we have: - lim cos(θ) = lim 1/cos(θ) = 1. - θ→0 - θ→0 - Hence, lim θ/sinθ = 1. - θ→0 - Then, lim sin(θ)/θ = 1. - θ→0 - Since lim tan(θ)/θ = lim(sin(θ)/cos(θ))/θ = lim sin(θ)/θ * lim 1/cosθ = 1*1 = 1. - θ→0 - θ→0 - θ→0 ### Example: - Prove that lim sin(x)/sin(bx) = 1/b. - x→0 - **Solution:** lim sin(x)/sin(bx) = lim (sin(x)/x) * (bx/sin(bx)) - x→0 - x→0 - = lim (sin(x)/x) * [b * (lim (bx)/sin(bx))] - x→0 - x→0 - = 1 * b/1 - = b ### Example: - Calculate lim (1-cos(2x))/x - x→0 - **Solution:** lim (1-cos(2x))/x = lim (1-cos(2x))/x * (2x/2x) - x→0 - x→0 - = lim (1-cos(2x))/(2x) * lim 2 - x→0 - x→0 - = 0 * 2 - = 0 ## Basic Limit Rules 1. **Powers:** lim x^n = a^n - x→00 2. **Exponentials:** lim a^x = 0, lim a^x = +∞. - x→-∞ - x→+∞ 3. **Composite functions:** If lim f(x) exists, then - x→a - lim f(x) = g(lim f(x)). - x→a - x→a 4. **Roots:** If n is an odd natural number and f(x) ≥ 0 in some neighborhood of x=a. Then lim √(n)(f(x)) = √(n)(lim f(x)). - x→a - x→a 5. **Products:** If lim f(x) = 0, and g(x) is bounded in a neighborhood of a, then lim f(x)g(x) = 0. - x→a - x→a 6. **Quotients:** If lim u(x) = 0, lim v(x) = +∞ and lim u(x)*(v(x))^2 = A, then lim [1 + (1/v(x)^2)] = A. - x→a - x→a - x→a - x→a ### Example: - Calculate lim[ (1/x) + 1/x^2]. - x→∞ - **Solution:** lim(1/x) = 0, lim(1/x^2) = 0, and lim x^2*[1/x*(1/x^2)] = lim(x/x^2) = lim (1/x) = 0. - x→∞ - x→∞ - x→∞ - x→∞ - x→∞ - Therefore, lim[ (1/x) + 1/x^2] = 0. - x→∞ ### Exercise: - Calculate the limit: lim sin(x^2)/x^2 - x→0 ### Example: - Calculate the limit: lim (1 + 1/x)^x - x→∞ - **Solution:** Let y = (1 + 1/x)^x. Taking log of both sides, we get: - ln(y) = ln[(1 + 1/x)^x] - = x*ln(1 + 1/x) - So, we can write the limit as: - lim y = lim e^[x*ln(1 + 1/x)] - x→∞ - x→∞ - = e^[lim x * ln(1 + 1/x)] - x→∞ - = e^[lim ln(1 + 1/x)/(1/x)] - x→∞ - = e^1 - = e ### Exercise: - Calculate the limit: lim (1 + 3x)^(1/x) - x→0 ### Example: - Calculate the limit: lim [(x^2 + 2x + 3)(2x+3)/(x^2 + 4)] - x→∞ - **Solution:** lim [(x^2 + 2x + 3)(2x+3)/(x^2 + 4)] = lim [(1 + 2/x + 3/x^2)(2 + 3/x)/(1 + 4/x^2)] - x→∞ - x→∞ - = (1 + 0 + 0)(2 + 0)/(1 + 0) - = 2 ## Continuity - A function f is said to be **continuous at a point x=a** if the following three conditions hold: - f(a) is defined. - lim f(x) exists. - x→a - lim f(x) = f(a). - x→a - If any of these conditions fail, then the function is said to be **discontinuous at x=a**. ### Note: - The function must be defined at the point a. - The function must have a limit at the point a. - The value of the function at the point a must be equal to the limit. ### Definition: - If f:A→ℝ is a function, then f is **continuous in A** if it is continuous at each point in A. ### Note: - It is important to remember that a function can be discontinuous at a point without being undefined at that point. This is often the case when the limit does not exist or when the limit exists but does not equal the value of the function at the point. ### Definition: - Let f:A→ℝ be a function and a∈A. If for every ε>0, there exists a δ>0 such that |x-a|<δ implies |f(x)-f(a)|<ε, then f is said to be **continuous at x=a**. ### Example: - Consider the function: - f(x) = x^2, x<0 - f(x) = 0, x=0 - f(x) = x+1, x>0 - Investigate the continuity of this function at x=0. - **Solution:** - f(0) is defined and is equal to 0. - lim f(x) = lim x^2 = 0. And, lim f(x) = lim (x + 1) = 1. - x→0- - x→0- - x→0+ - x→0+ - So, the limit does not exist (does not exist from both sides), therefore the function is discontinuous at x=0. ### Definition: - A function f is said to be **continuous from the right at x=a** if lim f(x) = f(a). - x→a+ - A function f is said to be **continuous from the left at x=a** if lim f(x) = f(a). - x→a- ### Property: - A function is continuous at a point if and only if it is continuous both from the right and from the left at that point. ### Example: - Consider the function: - f(x) = x^2 + 1, x≥1 - f(x) = 2x, x<1 - Investigate the continuity of this function at x=1. - **Solution:** - f(1) is defined and it is equal to 1^2 + 1 = 2 - lim f(x) = lim (x^2 + 1) = 2. And, lim f(x) = lim 2x = 2. - x→1+ - x→1+ - x→1- - x→1- - Therefore, the limit exists, and it is equal to the value the function at the point, so the function is continuous at x=1 ### Note: - If f:A→B and g:B→C are continuous functions, then the composite function g∘f:A→C is also continuous. - Let A⊂ℝ, f: A→ℝ be a function and a∈A. If f(a)≠0 and g(x) is a function that is continuous at the point f(a), then the function g(f(x)) is also continuous at the point x=a. ### Example: - f(x) = x^3 + 3x and g(x) = 1/x are continuous functions. Check if the function defined as (g∘f)(x) is continuous at any point. - **Solution:** (g∘f)(x) = g(f(x)) = 1/x^3 + 3x - (g∘f)(x) is continuous at any point except 0, because the function f(x) is continuous and g(x) is continuous except for the point x=0, and the value of the function f(x) is 0 at this point. ## Properties of Continuous Functions - A continuous function f is **bounded** on a closed interval [a,b]. This means there exist numbers M and m such that m≤f(x)≤M for all x∈[a,b]. - Let a<c<b. If f is continuous on the interval [a, b] and f(a) ≠ f(b), then for every number d between f(a) and f(b) there is at least one point c in the interval [a,b] such that f(c) = d. This is known as the **intermediate value theorem**. - If f is continuous on a closed interval [a,b], then f attains its maximum and minimum values on that interval. This is known as the **extreme value theorem**. ## Exercises: - Prove that the equation x^3 + x - 1 = 0 has a root in the interval (0,1). - Find the maximum and minimum values of the function f(x) = x^3 - 3x^2 + 2 on the interval [0,3]. ### Local extrema - If for every x in the neighborhood of a point c, except for x=c, f(x) ≤ f(c), then the function f has a **local maximum** at a point c. - If for every x in the neighborhood of a point c, except for x=c, f(x) ≥ f(c), then the function f has a **local minimum** at a point c. ### Absolute (Global) Extrema - The point P is the absolute maximum if f(P) is greater than f(x) for every x in the domain of f. - The point P is the absolute minimum if f(P) is smaller than f(x) for every x in the domain of f. ### Theorem: - Let f: [a,b]→ℝ be a continuous function on the interval [a, b]. Then f attains its absolute maximum and absolute minimum values in [a,b]. - The absolute extrema of f are attained at a point, a, b, or at critical points of f. Critical points of f are that points, c, such that either f’(c)=0, or f’(c) does not exist. ### Theorem: - If f:[a,b]→ℝ is a continuous function and f(a)≠f(b), then there exist some points c1,...,cn in the interval (a,b) such that f(a), f(c1), f(c2), ..., f(cn), f(b) are local extrema of f, where f(a) is the smallest value and f(b) is the largest value. - This means, that the absolute maximum and minimum values of the function are attained at the endpoints of the interval or at local extrema points. ### Theorem: - If f: [a,b]→ℝ is a continuous and strictly increasing function, then f:[a,b]→[f(a), f(b)] and its inverse function f⁻¹:[f(a), f(b)]→[a,b] are: - (i) Strictly increasing - (ii) Continuous