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APEX Calculus: for University of Lethbridge

Section 1.3 Finding Limits Analytically

In Section 1.1 we explored the concept of the limit without a strict definition, meaning we could only make approximations. In the previous section we gave the definition of the limit and demonstrated how to use it to verify our approximations were correct. Thus far, our method of finding a limit is
  1. make a really good approximation either graphically or numerically, and
  2. verify our approximation is correct using a \(\varepsilon\)-\(\delta\) proof.
Recognizing that \(\varepsilon\)-\(\delta\) proofs are cumbersome, this section gives a series of theorems which allow us to find limits much more quickly and intuitively.
Suppose that \(\lim_{x\to 2} f(x)=2\) and \(\lim_{x\to 2} g(x) = 3\text{.}\) What is \(\lim_{x\to 2}(f(x)+g(x))\text{?}\) Intuition tells us that the limit should be \(5\text{,}\) as we expect limits to behave in a nice way. The following theorem states that already established limits do behave nicely.
Figure 1.3.2. Video presentation of Theorem 1.3.1 (three videos)
We apply the theorem to an example.

Example 1.3.3. Using basic limit properties.

\begin{align*} \lim_{x\to 2} f(x)\amp=2\amp\lim_{x\to 2} g(x)\amp= 3\amp p(x)\amp = 3x^2-5x+7\text{.} \end{align*}
Find the following limits:
  1. \(\displaystyle \lim\limits_{x\to 2}(f(x) + g(x))\)
  2. \(\displaystyle \lim\limits_{x\to 2}(5f(x) + g(x)^2)\)
  3. \(\displaystyle \lim\limits_{x\to 2}p(x)\)
Solution 1. Video solution
Solution 2.
  1. Using the Sums/Differences property, we know that
    \begin{align*} \lim_{x\to 2}(f(x) + g(x)) \amp = \lim_{x\to 2}f(x) + \lim_{x\to 2}g(x)\\ \amp = 2+3 =5\text{.} \end{align*}
  2. Using the Scalar Multiples, Sums/Differences, and Powers properties, we find that
    \begin{align*} \lim_{x\to 2}(5f(x) + g(x)^2) \amp = \lim_{x\to 2}(5f(x))+\lim_{x\to 2}(g(x)^2)\\ \amp = 5\lim_{x\to 2}f(x) + \mathopen{}\left(\lim_{x\to 2}g(x)\right)^2\mathclose{}\\ \amp = 5\cdot 2 + 3^2 = 19\text{.} \end{align*}
  3. Here we combine the Powers, Scalar Multiples, Sums/Differences and Constants properties. We show quite a few steps, but in general these can be omitted:
    \begin{align*} \lim_{x\to 2} p(x) \amp = \lim_{x\to 2}\left(3x^2-5x+7\right)\\ \amp = \lim_{x\to 2}\mathopen{}\left(3x^2\right)\mathclose{}-\lim_{x\to 2}(5x)+\lim_{x\to 2}7\\ \amp = 3\bigl(\lim_{x\to 2}x\bigr)^2-5\lim_{x\to 2}(x) +7\\ \amp = 3\cdot 2^2 - 5\cdot 2+7\\ \amp = 9 \end{align*}
Part c of the previous example demonstrates how the limit of a quadratic polynomial can be determined using the properties of Theorem 1.3.1. Not only that, recognize that
\begin{equation*} \lim_{x\to 2} p(x) = 9 = p(2); \end{equation*}
i.e., the limit at \(2\) could have been found just by plugging \(2\) into the function. This holds true for all polynomials, and also for rational functions (which are quotients of polynomials), as stated in the following theorem.
Figure 1.3.5. Video presentation of Theorem 1.3.4 and Theorem 1.3.20

Example 1.3.6. Finding a limit of a rational function.

Using Theorem 1.3.4, find
\begin{equation*} \lim_{x\to -1} \frac{3x^2-5x+1}{x^4-x^2+3}\text{.} \end{equation*}
Using Theorem 1.3.4, we can quickly state that
\begin{align*} \lim_{x\to -1}\frac{3x^2-5x+1}{x^4-x^2+3} \amp = \frac{3(-1)^2-5(-1)+1}{(-1)^4-(-1)^2+3}\\ \amp = \frac{9}{3} =3\text{.} \end{align*}
It was likely frustrating in Section 1.2 to do a lot of work with \(\varepsilon\) and \(\delta\) to prove that
\begin{equation*} \lim_{x\to 2} x^2 = 4 \end{equation*}
as it seemed fairly obvious. The previous theorems state that many functions behave in such an “obvious” fashion, as demonstrated by the rational function in Example 1.3.6.
Polynomial and rational functions are not the only functions to behave in such a predictable way. The following theorem gives a list of functions whose behavior is particularly “nice” in terms of limits. In Section 1.5, we will give a formal name to these functions that behave “nicely.”

Example 1.3.8. Evaluating limits analytically.

Evaluate the following limits.
  1. \(\displaystyle \lim\limits_{x\to \pi} \cos(x)\)
  2. \(\displaystyle \lim\limits_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right)\)
  3. \(\displaystyle \lim\limits_{x\to \pi/2}(\cos(x)\sin(x))\)
  4. \(\displaystyle \lim\limits_{x\to 1} e^{\ln(x)}\)
  5. \(\displaystyle \lim\limits_{x\to 0} \dfrac{\sin(x)}{x}\)
  1. This is a straightforward application of Theorem 1.3.7: \(\lim\limits_{x\to \pi} \cos(x) = \cos(\pi) = -1\text{.}\)
  2. We can approach this in at least two ways. First, by directly applying Theorem 1.3.7, we have:
    \begin{equation*} \lim_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right) = \sec^2(3)-\tan^2(3)\text{.} \end{equation*}
    Using the Pythagorean Theorem, this last expression is \(1\text{;}\) therefore
    \begin{equation*} \lim_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right) = 1\text{.} \end{equation*}
    We can also use the Pythagorean Theorem from the start.
    \begin{equation*} \lim_{x\to 3} \left(\sec^2(x) - \tan^2(x)\right) = \lim_{x\to 3} 1 = 1\text{,} \end{equation*}
    using the Constants rule. Either way, we find the limit is \(1\text{.}\)
  3. Applying the Products rule and Theorem 1.3.7 gives
    \begin{equation*} \lim\limits_{x\to \pi/2} \cos(x)\sin(x) = \cos(\pi/2)\sin(\pi/2) = 0\cdot 1 = 0\text{.} \end{equation*}
  4. Again, we can approach this in two ways. First, we can use the exponential/logarithmic identity that \(e^{\ln(x)} = x\) and evaluate \(\lim\limits_{x\to 1} e^{\ln(x)} = \lim\limits_{x\to 1} x = 1\text{.}\)
    We can also use the Compositions rule. Using Theorem 1.3.7, we have \(\lim\limits_{x\to 1}\ln(x) = \ln(1) = 0\) and \(\lim_{x\to 0} e^x= e^0=1\text{,}\) satisfying the conditions of the Compositions rule. Applying this rule,
    \begin{equation*} \lim_{x\to 1} e^{\ln(x)} = e^{\lim_{x\to 1} \ln(x)}=e^{\ln(1)} = e^0 = 1\text{.} \end{equation*}
    Both approaches are valid, giving the same result.
  5. We encountered this limit in Section 1.1. Applying our theorems, we attempt to find the limit as
    \begin{equation*} \lim_{x\to 0}\frac{\sin(x)}{x}\rightarrow \frac{\sin(0) }{0}\text{,} \end{equation*}
    which is of the form \(\frac{0}{0}\text{.}\) This, of course, violates a condition of the Quotients rule, as the limit of the denominator is not allowed to be \(0\text{.}\) Therefore, we are still unable to evaluate this limit with tools we currently have at hand.
Based on what we’ve done so far, this section could have been titled “Using Known Limits to Find Unknown Limits.” By knowing certain limits of functions, we can find limits involving sums, products, powers, etc., of these functions. We further the development of such comparative tools with the Squeeze Theorem, a clever and intuitive way to find the value of some limits.
Before stating this theorem formally, suppose we have functions \(f\text{,}\) \(g\text{,}\) and \(h\) where \(g\) always takes on values between \(f\) and \(h\text{;}\) that is, for all \(x\) in an interval,
\begin{equation*} f(x) \leq g(x) \leq h(x)\text{.} \end{equation*}
If \(f\) and \(h\) have the same limit at \(c\text{,}\) and \(g\) is always “squeezed” between them, then \(g\) must have the same limit as well. That is what the Squeeze Theorem states. This is illustrated in Figure 1.3.9.
Graph that illustrates the squeeze theorem using functions h, g, and f.
Graph illustrating the squeeze theorem. There are three functions, \(h(x)\text{,}\) \(g(x)\text{,}\) and \(f(x)\text{.}\) On the \(y\)axis, there is a marker at \(y = 4\text{,}\) labeled \(L\) and on the \(x\)axis there is a marker at \(x = 5\text{,}\) labeled \(c\text{.}\)
For all values of \(x \leq c\) \(f(x) \leq L\) and \(h(x) \geq L\text{.}\) For all values of \(x\) \(f(x) \leq g(x) \leq h(x)\text{,}\) that is, the line of the function \(g(x)\) is between the lines of the functions \(g(x)\) and \(f(x)\text{.}\)
The graph shows that where \(x = c\text{,}\) \(f(x)\) and \(h(x)\) converge on \(y = L = 4\text{.}\) Because \(f(x) \leq g(x) \leq h(x)\text{,}\) we can extrapolate that \(\lim\limits_{x\to \c} g(x) = L\) too.
Figure 1.3.9. An illustration of the Squeeze Theorem
Figure 1.3.11. Explaining the Squeeze Theorem
It can take some work to figure out appropriate functions by which to “squeeze” a given function. However, that is generally the only place where work is necessary; the theorem makes the “evaluating the limit part” very simple.
The Squeeze Theorem can be used to show that limits of \(\sin(x)\) can be done by direct substitution, as the videos in Figure 1.3.12 illustrate.
Figure 1.3.12. Using the Squeeze Theorem to take the limit of \(\sin(x)\) at \(0\)
We use the Squeeze Theorem in the following example to finally prove that \(\lim\limits_{x\to 0} \frac{\sin(x)}{x} = 1\text{.}\)

Example 1.3.13. Using the Squeeze Theorem.

Use the Squeeze Theorem to show that
\begin{equation*} \lim_{x\to 0} \frac{\sin(x)}{x} = 1\text{.} \end{equation*}
Solution 1. Video solution
Solution 2.
We begin by considering the unit circle. Each point on the unit circle has coordinates \((\cos(\theta),\sin(\theta))\) for some angle \(\theta\) as shown in Figure 1.3.14. Using similar triangles, we can extend the line from the origin through the point to the point \((1,\tan(\theta))\text{,}\) as shown. (Here we are assuming that \(0\leq \theta \leq \pi/2\text{.}\) Later we will show that we can also consider \(\theta \leq 0\text{.}\))
Illustration of a unit Circle, with a right and acute Triangle in quadrant one.
Illustration of a unit circle, with a right and acute triangle in quadrant one. The right triangle purtudes out of the unit circle, while the acute triangle stays within the bounds of the circle. The acute triangle is also contained within the right triangle. There are horizontal lines and vertical lines inside the circle, these lines act only as guides. The horizontal and vertical lines intersect at \((0, 0)\text{.}\) The horizontal line starts at \((-1, 0)\) and ends at \(1, 0\text{,}\) the vertical line starts at \((0, 1)\) and ends at \((0, -1)\text{.}\)
The right Triangle is made up of the points \((0, 0)\text{,}\) \((1, \tan(\theta))\text{,}\) and \((1, 0)\text{.}\) The acute Triangle is made of the points \((0, 0)\text{,}\) \((\cos(\theta), \sin(\theta))\text{,}\) and \((1, 0)\text{.}\) The angle at point \((0, 0)\) is labeled as \(\theta\)
Figure 1.3.14. The unit circle and related triangles
Figure 1.3.14 shows three regions have been constructed in the first quadrant, two triangles and a sector of a circle, which are also drawn below. The area of the large triangle is \(\frac{1}{2}\tan(\theta)\text{;}\) the area of the sector is \(\theta/2\text{;}\) the area of the triangle contained inside the sector is \(\frac{1}{2}\sin(\theta)\text{.}\) It is then clear from Figure 1.3.15 that
\begin{equation*} \frac{\tan(\theta)}{2}\geq\frac{\theta}{2}\geq\frac{\sin(\theta)}{2}\text{.} \end{equation*}
(You may need to recall that the area of a sector of a circle is \(\frac{1}{2}r^2 \theta\) with \(\theta\) measured in radians.)
right triangle housing an acute triangle and sector of a circle.
A right triangle with two dotted lines inside it. One is curved and represents a sector of a circle inside the triangle. The sector of a circle has an area of \(\frac{1}{2}r^2 \theta\text{.}\) The other dotted line represents an acute triangle housed within the larger right triangle.
The right triangle has the side lengths 1 for its base and \(\tan(\theta)\) for the vertical line. Where these lines connect is the right angle. There is an angle \(\theta\) that is shared by both triangles and the circle sector.
Sector of a circle from previous image with dotted lines creating the right and acute triangles.
Sector of a circle from Figure 1.3.15.(a) with dotted lines creating the right triangle from Figure 1.3.15.(a), there is also a dotted line showing the same acute triangle from Figure 1.3.15.(a). The acute triangle is contained inside the circle sector, while the right triangle purtrudes out of it.
There is an angle \(\theta\) that is shared by both triangles and the circle sector. They also share the length 1 for the horizontal line.
Acute triangle from previous image with dotted lines creating the two right triangle an circle sector.
Acute triangle from Figure 1.3.15.(a) with dotted lines representing the circle sector from Figure 1.3.15.(b) and right triangle from Figure 1.3.15.(a). There is an additional dotted line that creates a second right triangle within the acute triangle. Both the sector of a circle and right triangle fall outside the bounds of the acute triangle.
The second right triangle a height of \(\sin(\theta)\text{.}\) As with Figure 1.3.15.(a) and Figure 1.3.15.(b) the horizontal line of the triangle is \(1\text{.}\)
Figure 1.3.15. Bounding the sector between two triangles
Multiply all terms by \(\frac{2}{\sin(\theta)}\text{,}\) giving
\begin{equation*} \frac{1}{\cos(\theta)} \geq \frac{\theta}{\sin(\theta)} \geq 1\text{.} \end{equation*}
Taking reciprocals reverses the inequalities, giving
\begin{equation*} \cos(\theta) \leq \frac{\sin(\theta)}{\theta} \leq 1\text{.} \end{equation*}
(These inequalities hold for all values of \(\theta\) near \(0\text{,}\) even negative values, since \(\cos(-\theta) = \cos(\theta)\) and \(\sin(-\theta) = -\sin(\theta)\text{.}\))
Now take limits.
\begin{align*} \lim_{\theta\to 0} \cos(\theta)\amp \leq \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} \leq \lim_{\theta\to 0} 1\\ \cos(0) \amp\leq \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} \leq 1\\ 1\amp \leq \lim_{\theta\to 0} \frac{\sin(\theta)}{\theta} \leq 1 \end{align*}
Clearly this means that \(\lim\limits_{\theta\to 0} \frac{\sin(\theta)}{\theta}=1\text{.}\)
With the limit \(\lim\limits_{\theta\to 0} \frac{\sin(\theta)}{\theta}=1\) finally established, we can move on to other limits involving trigonometric functions, as the video in Figure 1.3.16 demonstrates.
Figure 1.3.16. Finding limits involving trigonometric functions
Two notes about the Example 1.3.13 are worth mentioning. First, one might be discouraged by this application, thinking “I would never have come up with that on my own. This is too hard!” Don’t be discouraged; within this text we will guide you in your use of the Squeeze Theorem. As one gains mathematical maturity, clever proofs like this are easier and easier to create.
Second, this limit tells us more than just that as \(x\) approaches \(0\text{,}\) \(\sin(x)/x\) approaches \(1\text{.}\) Both \(x\) and \(\sin(x)\) are approaching \(0\text{,}\) but the ratio of \(x\) and \(\sin(x)\) approaches \(1\text{,}\) meaning that they are approaching \(0\) in essentially the same way. Another way of viewing this is: for small \(x\text{,}\) the functions \(y=x\) and \(y=\sin(x)\) are essentially indistinguishable.
We include this special limit, along with three others, in the following theorem.
A short word on how to interpret the latter three limits. We know that as \(x\) goes to \(0\text{,}\) \(\cos(x)\) goes to \(1\text{.}\) So, in the second limit, both the numerator and denominator are approaching \(0\text{.}\) However, since the limit is \(0\text{,}\) we can interpret this as saying that “\(\cos(x)\) is approaching \(1\) faster than \(x\) is approaching \(0\text{.}\)
In the third limit, inside the parentheses we have an expression that is approaching \(1\) (though never equaling \(1\)), and we know that \(1\) raised to any power is still \(1\text{.}\) At the same time, the power is growing toward infinity. What happens to a number near \(1\) raised to a very large power? In this particular case, the result approaches Euler’s number, \(e\text{,}\) approximately \(2.718\text{.}\)
In the fourth limit, we see that as \(x\to 0\text{,}\) \(e^x\) approaches \(1\) “just as fast” as \(x\to 0\text{,}\) resulting in a limit of \(1\text{.}\)
The special limits stated in Theorem 1.3.17 are called indeterminate forms; in this case they are of the form \(0/0\text{,}\) except the third limit, which is of a different form. You’ll learn techniques to find these limits exactly using calculus in Section 6.7.
Our final theorem for this section will be motivated by the following example.

Example 1.3.18. Using algebra to evaluate a limit.

Evaluate the following limit:
\begin{equation*} \lim_{x\to 1}\frac{x^2-1}{x-1}\text{.} \end{equation*}
We begin by attempting to apply Theorem 1.3.4 and substituting \(1\) for \(x\) in the quotient. This gives:
\begin{equation*} \lim_{x\to 1}\frac{x^2-1}{x-1} = \frac{1^2-1}{1-1} \end{equation*}
which is of the form \(\frac{0}{0}\text{,}\) an indeterminate form. We cannot apply the theorem.
By graphing the function, as in Figure 1.3.19, we see that the function seems to be linear, implying that the limit should be easy to evaluate. Recognize that the numerator of our quotient can be factored:
\begin{equation*} \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1}\text{.} \end{equation*}
The function is not defined when \(x=1\text{,}\) but for all other \(x\text{,}\)
\begin{align*} \frac{x^2-1}{x-1}\amp = \frac{(x-1)(x+1)}{x-1}\\ \amp = \frac{\cancel{(x-1)}(x+1)}{\cancel{(x-1)} }\\ \amp = x+1, \quad \text{if } x \neq 1 \end{align*}
Graph of the polynomial x squared minus 1 divided by the polynomial x minus 1. Is undefined at x = 1.
Graph of the linear equation \((x^1-1)/(x-1)\text{,}\) shows the function with the \(y\) interval \(0\) to \(3\) and \(x\) interval \(0\) to \(2\text{.}\) The function has a \(y\) intercept at \(y = 1\text{,}\) and is undefined at \(x = 1\text{.}\) The graph has a positive slope.
Figure 1.3.19. Graphing \(f\) in Example 1.3.18 to understand a limit
Clearly \(\lim\limits_{x\to 1}(x+1) = 2\text{.}\) Recall that when considering limits, we are not concerned with the value of the function at \(1\text{,}\) only the value the function approaches as \(x\) approaches \(1\text{.}\) Since \((x^2-1)/(x-1)\) and \(x+1\) are the same at all points except at \(x=1\text{,}\) they both approach the same value as \(x\) approaches \(1\text{.}\) Therefore we can conclude that
\begin{align*} \lim_{x \to 1} \frac{x^2-1}{x-1} \amp =\lim_{x \to 1} (x+1)\\ \amp =2 \end{align*}
The key to Example 1.3.18 is that the functions \(y=(x^2-1)/(x-1)\) and \(y=x+1\) are identical except at \(x=1\text{.}\) Since limits describe a value the function is approaching, not the value the function actually attains, the limits of the two functions are always equal.
The Fundamental Theorem of Algebra tells us that when dealing with a rational function of the form \(g(x)/f(x)\) and directly evaluating the limit \(\lim\limits_{x\to c} \frac{g(x)}{f(x)}\) returns “0/0”, then \((x-c)\) is a factor of both \(g(x)\) and \(f(x)\text{.}\) One can then use algebra to factor this binomial out, cancel, then apply Theorem 1.3.20. We demonstrate this once more.

Example 1.3.21. Evaluating a limit using Theorem 1.3.20.

\begin{equation*} \lim\limits_{x\to 3} \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15}\text{.} \end{equation*}
Solution 1. Video solution
Solution 2.
We attempt to apply Theorem 1.3.4 by substituting \(3\) for \(x\text{.}\) This returns the familiar indeterminate form of “0/0”. Since the numerator and denominator are each polynomials, we know that \((x-3)\) is factor of each. Using whatever method is most comfortable to you, factor out \((x-3)\) from each (using polynomial division, synthetic division, a computer algebra system, etc.). We find that
\begin{equation*} \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} = \frac{(x-3)\left(x^2+x-2\right)}{(x-3)\left(2 x^2+9 x-5\right)}\text{.} \end{equation*}
We can cancel the \((x-3)\) factors as long as \(x\neq 3\text{.}\) Using Theorem 1.3.20 we conclude:
\begin{align*} \lim_{x\to 3} \frac{x^3-2 x^2-5 x+6}{2 x^3+3 x^2-32 x+15} \amp = \lim_{x\to 3}\frac{(x-3)\left(x^2+x-2\right)}{(x-3)\left(2 x^2+9 x-5\right)}\\ \amp =\lim_{x\to 3} \frac{x^2+x-2}{2 x^2+9 x-5}\\ \amp = \frac{10}{40}\\ \amp = \frac{1}{4}\text{.} \end{align*}

Example 1.3.22. Evaluating a Limit with a Hole.

\begin{equation*} \lim\limits_{x\to 9} \frac{\sqrt{x}-3}{x-9}\text{.} \end{equation*}
Solution 1. Video solution
Solution 2.
We begin by trying to apply the Quotients limit rule, but the denominator evaluates to zero. In fact, this limit is of the indeterminate form \(0/0\text{.}\) We will do some algebra to resolve the indeterminate form. In this case, we multiply the numerator and denominator by the conjugate of the numerator.
\begin{align*} \frac{\sqrt{x}-3}{x-9} \amp = \frac{\sqrt{x}-3}{x-9} \cdot \frac{\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)}\\ \amp= \frac{x-9}{(x-9)(\sqrt{x}+3)} \end{align*}
We can cancel the \((x-9)\) factors as long as \(x\neq 9\text{.}\) Using Theorem 1.3.20 we conclude:
\begin{align*} \lim_{x\to 9} \frac{\sqrt{x}-3}{x-9} \amp =\lim_{x\to 9} \frac{x-9}{(x-9)\left(\sqrt{x}+3\right)}\\ \amp = \lim_{x\to 9 }\frac{1}{\sqrt{x}+3}\\ \amp = \frac{1}{\lim_{x\to 9}\sqrt{x}+\lim_{x \to 9}3}\\ \amp = \frac{1}{\sqrt{\lim_{x\to 9}x}+3}\\ \amp = \frac{1}{\sqrt{3+3}}\\ \amp = \frac{1}{6}\text{.} \end{align*}
We end this section by revisiting a limit first seen in Section 1.1, a limit of a difference quotient. Let \(f(x) = -1.5x^2+11.5x\text{;}\) we approximated the limit \(\lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}\approx 8.5\text{.}\) We formally evaluate this limit in the following example.

Example 1.3.23. Evaluating the limit of a difference quotient.

Let \(f(x) = -1.5x^2+11.5x\text{;}\) find \(\lim\limits_{h\to 0}\frac{f(1+h)-f(1)}{h}\text{.}\)
Since \(f\) is a polynomial, our first attempt should be to employ Theorem 1.3.4 and substitute \(0\) for \(h\text{.}\) However, we see that this gives us “\(0/0\text{.}\)” Knowing that we have a rational function hints that some algebra will help. Consider the following steps:
\begin{align*} \lim_{h\to 0}\frac{f(1+h)-f(1)}{h}\amp = \lim_{h\to 0}\frac{-1.5(1+h)^2 + 11.5(1+h) - \left(-1.5(1)^2+11.5(1)\right)}{h}\\ \amp =\lim_{h\to 0}\frac{-1.5(1+2h+h^2) + 11.5+11.5h - 10}{h}\\ \amp =\lim_{h\to 0}\frac{-1.5h^2 +8.5h}{h}\\ \amp =\lim_{h\to 0}\frac{h(-1.5h+8.5)}h\\ \amp =\lim_{h\to 0}(-1.5h+8.5) \quad (\text{using } \knowl{./knowl/xref/thm_limit_allbut1.html}{\text{Theorem 1.3.20}}, \text{ as } h\neq 0 )\\ \amp =8.5 \quad (\text{using } \knowl{./knowl/xref/thm_poly_rat.html}{\text{Theorem 1.3.4}} ) \end{align*}
This matches our previous approximation.
This section contains several valuable tools for evaluating limits. One of the main results of this section is Theorem 1.3.7; it states that many functions that we use regularly behave in a very nice, predictable way. In Section 1.5 we give a name to this nice behavior; we label such functions as continuous. Defining that term will require us to look again at what a limit is and what causes limits to not exist.

Exercises Exercises

Terms and Concepts

Explain in your own words, without using \(\varepsilon\)-\(\delta\) formality, why \(\lim\limits_{x\to c}b=b\text{.}\)
Explain in your own words, without using \(\varepsilon\)-\(\delta\) formality, why \(\lim\limits_{x\to c}x=c\text{.}\)
What does the text mean when it says that certain functions’ “behavior is ‘nice’ in terms of limits”? What, in particular, is “nice”?
Sketch a graph that visually demonstrates the Squeeze Theorem.
You are given the following information:
\begin{equation*} \begin{aligned} \lim_{x\to 1}f(x)\amp=0\amp\lim_{x\to 1}g(x)\amp=0\amp\lim_{x\to 1}\frac{f(x)}{g(x)}\amp=2 \end{aligned} \end{equation*}
What can be said about the relative sizes of \(f(x)\) and \(g(x)\) as \(x\) approaches \(1\text{?}\)
  • True
  • False
\(\lim\limits_{x\to 1}\ln x = 0\text{.}\)
Use a theorem to defend your answer.


Exercise Group.
Use the following information to evaluate the given limit, when possible.
\begin{align*} \lim\limits_{x\to9}f(x)\amp=6\amp\lim\limits_{x\to6}f(x)\amp=9\amp f(9)\amp=6\\ \lim\limits_{x\to9}g(x)\amp=3\amp\lim\limits_{x\to6}g(x)\amp=3\amp g(6)\amp=3 \end{align*}
\(\lim\limits_{x\to 9}(f(x)+g(x))\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 9}\left(\frac{3f(x)}{g(x)}\right)\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 9}\left(\frac{f(x)-2g(x)}{g(x)}\right)\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 6}\left(\frac{f(x)}{3-g(x)}\right)\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 9}g(f(x))\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 6}f(g(x))\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 6}g(f(f(x)))\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 6}\left(f(x)g(x)-f(x)^2+g(x)^2\right)\)
If it is not possible to determine the limit, state why not.
Exercise Group.
Use the following information to evaluate the given limit, when possible. If it is not possible to determine the limit, state why not.
\begin{align*} \lim_{x\to1}f(x)\amp=2\amp\lim_{x\to10}f(x)\amp=1\amp f(1)\amp=1/5\\ \lim_{x\to1}g(x)\amp=0\amp\lim_{x\to10}g(x)\amp=\pi\amp g(10)\amp=\pi \end{align*}
\(\lim\limits_{x\to 1}\left(f(x)g(x)\right)\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 10}\cos(g(x))\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 1}g(5f(x))\)
If it is not possible to determine the limit, state why not.
\(\lim\limits_{x\to 1}5^{g(x)}\)
If it is not possible to determine the limit, state why not.
Exercise Group.
Evaluate the given limit.
\(\lim\limits_{x\to 6}\left({x^{2}-3x+5}\right)\)
\(\lim\limits_{x\to {\frac{\pi }{6}}}\cos(x)\sin(x)\)
\(\lim\limits_{x\to 2}{4^{x^{3}-2x}}\)
\(\lim\limits_{x\to {\frac{\pi }{3}}}\csc(x)\)
\(\lim\limits_{x\to 5}{\frac{x^{2}-11x+30}{x^{2}-14x+45}}\)
\(\lim\limits_{x\to 9}{\frac{x^{2}-x-72}{x^{2}-14x+45}}\)
\(\lim\limits_{x\to -8}{\frac{x^{2}+3x-40}{x^{2}+13x+40}}\)
\(\lim\limits_{x\to -6}{\frac{x^{2}+8x+12}{x^{2}+3x-18}}\)
\(\lim\limits_{x\to -4}{\frac{x^{2}+13x+36}{x^{2}+12x+32}}\)
Exercise Group.
Use the Squeeze Theorem to evaluate the limit.
Show your work.
Show your work.
\(\lim\limits_{x\to1} f(x)\text{,}\) where \(3x-2\leq f(x) \leq x^3\)
Show your work.
\(\lim\limits_{x\to3} f(x)\text{,}\) where \(6x-9\leq f(x) \leq x^2\)
Show your work.
Exercise Group.
The following exercises challenge your understanding of limits but can be evaluated using the knowledge gained in Section 1.3.
\(\lim\limits_{x\to0}\frac{\sin(x)}{x}\text{,}\) where \(x\) is measured in degrees, not radians.
Let \(f(x)=0\) and \(g(x)=\frac{x}{x}\text{.}\)
Explain why \(\lim\limits_{x\to2}f(x)=0\text{.}\)
Explain why \(\lim\limits_{x\to0}g(x)=1\text{.}\)
Explain why \(\lim\limits_{x\to2} g(f(x))\) does not exist.
Explain why the previous statement does not violate the Composition Rule of Theorem 1.3.1.