Lecture Notes – Week 4 PDF

## Limit First, we will talk about the motivation of doing limit, it has many applications itself, but it is also used in derivatives and integrals. ### Average and Instant Velocity We talked about the position function of an object thrown vertically. If a rock is launched vertically upward wit...

## Limit First, we will talk about the motivation of doing limit, it has many applications itself, but it is also used in derivatives and integrals. ### Average and Instant Velocity We talked about the position function of an object thrown vertically. If a rock is launched vertically upward with a speed of 96 m/s, the position of this rock after *t* seconds is given by *h(t) = -16t^2 + 96t*. Recall that *h(0) = 50 m*, and we can find the maximum height by *h’(t) = h(3)*, which is 144 m. So, the rock in 3s goes to 144m and then it comes down. Now, let’s find the average speed of this rock for different intervals: * (a) *t₁ = 1s* and *t₂= 2s* * (b) *t₁ = 1s* and *t₃ = 3s* In general, the average speed of an object with the position function *s(t)* is obtained by *V_av = (s(t₂) - s(t₁)) / (t₂ - t₁)* Which is the slope of the line between the position of the object at *t = t₁* and the one at *t = t₂*. * (a) V_av = (h(t₂) - h(t₁)) / (t₂ - t₁) = (128 - 80) / (2 - 1) = 48 m/s * (b) V_av = (h(t₃) - h(t₁)) / (t₃ - t₁) = (144 - 80) / (3 - 1) = 32 m/s Let’s look at these value in the graph of *h(t)*. (A line joining two points on a curve is called a secant line) As we expected the slope of the line from 1 to 2 is more than the slope of the line from 1 to 3. The question is that how can we find the instant velocity of the rock at any given time, say *t = 1*? To find the instant speed, we choose times closer and closer to *t = 1*. If we calculate the average speed for those, we’ll see that they are getting closer and closer to a value, in this case, 64 m/s. | Time Interval | Average Velocity | |:---|:---| |[1, 2] | 48| |[1, 1.5] | 56 | |[1, 1.1] | 62.4 | |[1, 1.01] | 63.84 | |[1, 1.001] | 63.984| |[1, 1.0001] | 63.9984 | In fact, we find the limit of average speed, when the time approaches *t = 1*: *v(1) = lim (h(t) - h(1)) / (t - 1) = lim (-16t^2 + 96t -80) / (t - 1) = 64 m/s.* We will study the method to find this limit later. ### Definitions **Definition:** Suppose the function *f* is defined for all *x* near *a*, except possibly at *a*. If *f(x)* is arbitrary close to *L* for all *x* sufficiently close to *a* (but not equal), we write *lim*_{*x → a*} *f(x)* = *L* and say the *limit of f(x) as x approaches a is L*. **Example:** We use the graph of *f* to determine the following values. * Find *f(1)*, *f(2)*, and *f(3)*. * lim*_*x→1* *f(x)* = ? As you see from the graph, *x* approaches 1, the values for *f(x)* are getting closer and closer to 1.50 *lim*_*x→1* *f(x)* = 1 *lim*_*x→2* *f(x)* = ? As you see, the values of *f(x)* in this case are getting to 2, even *f(2)* is not 2! So, *lim*_*x→2* *f(x)* = 2 *lim*_*x→3* *f(x)* = ? *f* is not defined at *x = 3*, but still the limit exists, by the same argument: *lim*_*x→3* *f(x)* = 3. **Example:** Let’s use a table now. We create a table of values of *f(x) = √x-1 / x-1* for the values near 1. We will use it to find *lim*_*x→1* *f(x)*. | x | f(x) = √x-1/x-1 | |:---:|:---| | 0.9 | 0.5131670 | | 0.99 | 0.5012563 | | 0.999 | 0.5001251 | | 0.9999 | 0.5000125 | | 1.0001 | 0.4999875| | 1.001 | 0.4998750 | | 1.01 | 0.4987562| | 1.1 | 0.4880885 | As you see in the table, when *x* approaches to 1, *f(x)* is getting closer and closer to 0.5, so *lim*_*x→1* *f(x)* = 0.5 (= 1/2) In this table, we can see values of *f(x)* when *x* approaches to 1 from higher/lower values. In this example both are getting to 0.5. Also, note that *x = 1* is not in the domain of *f*, but we have the *lim*_*x→1**f*. ### One-Sided Limits **Definition:** Suppose *f* is defined for all *x* near *a* with *x > a*. If *f(x)* is arbitrary close to *L* for all *x* sufficiently close to *a* with *x > a*, we write *lim*_*x→a+* *f(x)* = *L* and say the *limit of f(x) as x approaches a from the right equals L*. *lim*_*x→a-* *f(x)* = *L* and say the *limit of f(x) as x approaches a from the left equals L*. **Example:** Let *f(x) = x^2 - 8 / 4x-2*. We want to find *lim*_*x→2+* *f(x)*, *lim*_*x→2-* *f(x)*, and *lim*_*x→2* *f(x)* if they exist. The graph of *f* is given, note that *2 ∉ D_f* It’s clear (from its graph) that if *x* approaches 2 from right, *f(x)* is getting to 3, same for approaching from left, and finally when *x > 2*, then *f(x) → 3*. So, *lim*_*x→2+* *f(x)* = 3 *lim*_*x→2-* *f(x)* = 3 *lim*_*x→2* *f(x)* = 3 We can also find these limits from the value table of *f*. | x | f(x) = (x^2 - 8) / (4x - 2) | |:---|:---| | 1.9 | 2.8525 | | 1.99 | 2.985025 | | 1.999 | 2.99850025 | | 1.9999 | 2.99985000 | | 2.0001 | 3.00015000 | | 2.001 | 3.00150025 | | 2.01 | 3.015025 | | 2.1 | 3.1525 | As you see in this example the right and left limits exist and they are equal, and the *lim*_*x→2* *f(x)* also exists and equals them. ### Theorem **Theorem:** Assume *f* is defined for all *x* near *a*, except possibly at *a*. Then *lim*_*x→a* *f(x)* = *L* if and only if *lim*_*x→a+* *f(x)* = *L* and *lim*_*x→a-* *f(x)* = *L* **Example:** Given the graph of *g*, find the following limits, if they exist. * lim*_*x→2-* *g(x)* = 4 (*x < 2*) * lim*_*x→2+* *g(x)* = 2 (*x > 2*) So, *lim*_*x→2* *g(x)* does not exist. **Example:** Consider the function *f(x) = cos(1/x)*. For any point *a*, except 0, the *lim*_*x→a* *f(x)* exists but for *a = 0*, as you see from its graph, the right limit does not exist, because the graph is alternatively changes between -1 and 1 and does not approach to one value, so *lim*_*x→0+* *cos(1/x)* does not exist, so *lim*_*x→0* *cos(1/x)* does not exist (same for *lim*_*x→0-* *f(x)*). We have the same argument for *g(x) = sin(1/x)*. ### Techniques for Computing Limits 1. **Let’s start with a constant function. If *f* is a constant function, *f(x) = b*, for some *b ∈ R*, then it’s clear that *lim*_*x→a* *f(x)* = *lim*_*x→a* *b* = *b*. 2. **For a linear function *y = mx + b*.** As an example, we want to find *lim*_*x→1* *(x + 1)*, as you can see, approaching *x* to 1, *f(x)* is getting to 2, so *lim*_*x→1* *f(x)* = 2. But what is 2? Isn’t it *f(1)*? This is the case for any value of *a*, for any linear function. *lim*_*x→a* *f(x)* = *lim*_*x→a* *(mx + b)* = *ma + b* = *f(a).* Before continue to other functions, let’s have the rules for limit. **Theorem:** Assume *lim*_*x→a**f(x)* and *lim*_*x→a**g(x)* exist. For any real number *c*, and positive integers *m* and *n*, we have * (i) *lim*_*x→a* *(f(x) + g(x))* = *lim*_*x→a* *f(x)* + *lim*_*x→a* *g(x)* * (ii) *lim*_*x→a* *(c * f(x))* = *c* *lim*_*x→a* *f(x)* * (iii) *lim*_*x→a* *(f(x) - g(x))* = *lim*_*x→a* *f(x)* - *lim*_*x→a* *g(x)* * (iv) *lim*_*x→a* *(f(x) / g(x))* = *lim*_*x→a**f(x)* / *lim*_*x→a* *g(x)* (*lim*_*x→a* *g(x) ≠ 0*) * (v) *lim*_*x→a* *(f(x))^n* = (*lim*_*x→a* *f(x)*)^n* * (vi) *lim*_*x→a* *(f(x))^m/n* = (*lim*_*x→a* *f(x)*)^m/n* (*n ≠ 0* if *n* is even) Note that (i), and (iii) hold for any number of functions. Also, note that in all of them *x ≠ a*. Lastly, all these laws are valid for one-sided limits. 3. **Let’s have an example for a polynomial:** *lim*_*x→1* *(3x^3 + 7x^2 - 2x + 1)* = *lim*_*x→1* *(3x^3)* + *lim*_*x→1**(7x^2)* + *lim*_*x→1**(-2x)* + *lim*_*x→1**(1)* = 3 *lim*_*x→1* *(x^3)* + 7 *lim*_*x→1* *(x^2)* + (-2) *lim*_*x→1* *(x)* + 1 = 3 (*lim*_*x→1* *x*)^3 + 7 (*lim*_*x→1* *x*)^2 - 2 *lim*_*x→1* *x* + 1 = 3 *1^3* + 7 *1^2* - 2 *1* + 1 = 9 = *f(1)* So, for any given polynomial *f(x)*: *lim*_*x→b* *f(x)* = *lim*_*x→b* *(anx^n + a_n-1x^n-1 + … + aix + a₀)* = *anb^n + a_n-1b_n-1 + … + a₁b + a₀* = *f(b)* 4. **A rational function is of the form *f(x) = p(x) / q(x)* where *p(x)* and *q(x)* are polynomials, so *lim*_*x→a* *f(x)* = *lim*_*x→a* *p(x)* / *lim*_*x→a* *q(x)* = *p(a) / q(a)* (*= f(a)*) provided that *lim*_*x→a* *q(x) ≠ 0*, and *q(a) ≠ 0*. **Example:** Suppose that *lim*_*x→2* *f(x)* = 4, *lim*_*x→2* *g(x)* = 5, and *lim*_*x→2* *h(x)* = 8, then * (a) *lim*_*x→2* *(f(x) - g(x)) / h(x)* = *lim*_*x→2* *(f(x) - g(x)) / *lim*_*x→2* *h(x)* = *(lim*_*x→2* *f(x) - lim*_*x→2* *g(x)) / *lim*_*x→2* *h(x)* = (4 - 5) / 8 = -1 / 8 * (b) *lim*_*x→2* *(6 * f(x) *g(x)) - h(x)* = *lim*_*x→2* *(6 * f(x) *g(x))* - *lim*_*x→2* *h(x)* = 6 *lim*_*x→2* *f(x)* *lim*_*x→2* *g(x)* - *lim*_*x→2* *h(x)* = 6 * 4 * 5 + 8 = 128 **Example:** If *f(x) = 3x^2 - 4x / 5x^2 - 36* then *lim*_*x→2* *f(x)* = *lim*_*x→2* *(3x^2 - 4x) / (5x^2 - 36)* = *(3 * 2^2 - 4 * 2) / (5 * 2^2 - 36)* = 4 / 1 **Example:** Find *lim*_*x→1* *(5x^4 - 3x^2 + 8x - 6) / (x + 1)* 5. **Alg functions.** Let's do it by an example. **Example:** We will calculate the following limit. *lim*_*x→2* *√(2x^2 + 9) + 3x - 1 / 4x + 1* = *lim*_*x→2* *(√(2x^2 + 9) + 3x - 1) / *lim*_*x→2* *(4x + 1)* = (*lim*_*x→2* *(√(2x^2 + 9))½* + *lim*_*x→2* *(3x - 1)* ) / *lim*_*x→2* *(4x + 1)* = (√(2 * 2^2 + 9))½ + (3 * 2 - 1) / 4 * 2 + 1 = √(25) + 5 / 9 = 10 / 9 Note that the limit for *x < √(25/2)* does not exist (why?) and to *x = √(25/2)*, we can approach only from right (why? Find it!) For piecewise functions, we should be more careful for the boarders, as the left and the right limit may differ. **Example:** Let *f(x) = -2x + 4 if x ≤ 1* and *√(x - 1)* if *x > 1*. Find *lim*_*x→1-* *f(x)* = 0 *lim*_*x→1+* *f(x)* = 2 *lim*_*x→1**f(x)* does not exist! In a rational (or alg) function, if we have 0/0, we need to use other techniques. In these examples we see two of them. **Example:** (a) *lim*_*x→2* *(x^2 - 6x + 8) / (x^2 - 4)* = *lim*_*x→2**(x - 2)(x - 4) / (x - 2)(x + 2)* = *lim*_*x→2* *(x - 4) / (x + 2)* = (2 - 4) / (2 + 2) = -2 / 4 = -1 / 2 Note that the cancellation is allowed in the limit, not for the function itself, because *f(x) = (x^2 - 6x + 8) / (x^2 - 4)* is different from *g(x) = (x - 4) / (x + 2)* why? (b) *lim*_*x→1* *√x - 1 / x - 1* = *lim*_*x→1* *√x - 1 / x - 1* *lim*_*x→1* *(√x + 1) / (√x + 1)* = *lim*_*x→1* *(x - 1) / (x - 1) * (√x + 1)* = *lim*_*x→1* *(√x + 1)* = √1 + 1 = 2 **Example:** Find *lim*_*x→5* *(x^2 - 7x + 10) / (x - 5)* ### Theorem: **Theorem:** Assume the functions *f, g*, and *h* satisfy *f(x) ≤ g(x) ≤ h(x)* for all values of *x* near *a*, except possibly at *a*. If *lim*_*x→a* *f(x) = lim*_*x→a* *h(x) = L*, then *lim*_*x→a* *g(x) = L*. We call this theorem, the squeeze theorem. (*g* is squeezed between *f* and *h* near *a*) **Example:** For -π/2 ≤ x ≤ π/2, we can show that (a) |-1| ≤ sin x ≤ |x| and (b) 0 ≤ 1 - cos x ≤ |x|. (Try to prove these inequalities) Using squeeze theorem, we can show that *lim*_*x→0* *sin x* = 0 and *lim*_*x→0* *cos x* = 1 (Write the details) **Example:** Recall that *lim*_*x→0* *sin(1/x)* does not exist, but we can show that *lim*_*x→0* *x sin(1/x)* = 0. We know that for any *x* -1 ≤ sin(1/x) ≤ 1 (shown before) ⇒ -x^2 ≤ x^2 sin(1/x) ≤ x^2 And *lim*_*x→0* *(-x^2)* ≤ *lim*_*x→0* *x^2 sin(1/x)* ≤ *lim*_*x→0* *(x^2)* ⇒ *lim*_*x→o* *x^2 sin(1/x)* = 0 **Example:** Suppose that *f* satisfies: 1 ≤ *f(x)* ≤ 1 + x^2 for all values near zero. Find *lim*_*x→0* *f(x)*.