Skills Review for Vector-Valued Functions and Space Curves

Learning Outcomes

Evaluate trigonometric functions using the unit circle
Recognize the basic limit laws
Use the limit laws to evaluate the limit of a function
Recognize when to apply L’Hôpital’s rule

In the Vector-Valued Functions and Space Curves section, we will define and develop skills needed to understand and work with vector-valued functions. Here we will review how to evaluate trigonometric functions at specific angle measures, apply limit laws, and apply L’Hôpital’s rule.

Evaluate Trigonometric Functions Using the Unit Circle

(also in Module 1, Skills Review for Polar Coordinates)
The unit circle tells us the value of cosine and sine at any of the given angle measures seen below. The first coordinate in each ordered pair is the value of cosine at the given angle measure, while the second coordinate in each ordered pair is the value of sine at the given angle measure. You will learn that all trigonometric functions can be written in terms of sine and cosine. Thus, if you can evaluate sine and cosine at various angle values, you can also evaluate the other trigonometric functions at various angle values. Take time to learn the $\left(x,y\right)$ coordinates of all of the major angles in the first quadrant of the unit circle.

Remember, every angle in quadrant two, three, or four has a reference angle that lies in quadrant one. The quadrant of the original angle only affects the sign (positive or negative) of a trigonometric function’s value at a given angle.

f-d-43392176e093fa07f39e1f3687226d4d751b809be928c16abac5dcb3+IMAGE+IMAGE

A General Note: Evaluating Tangent, Secant, Cosecant, and Cotangent Functions

If $\theta$ is an angle measure, then, using the unit circle,

$\begin{gathered}\tan \theta=\frac{sin \theta}{cos \theta}\\ \sec \theta=\frac{1}{cos \theta}\\ \csc t=\frac{1}{sin \theta}\\ \cot \theta=\frac{cos \theta}{sin \theta}\end{gathered}$

Example: Using the Unit Circle to Find the Value of Trigonometric Functions

Find $\sin \theta,\cos \theta,\tan \theta,\sec \theta,\csc \theta$ , and $\cot \theta$ when $\theta=\frac{\pi }{3}$ .

Show Solution

$\begin{align}&\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}\\ &\cos \frac{\pi }{3}=\frac{1}{2}\\ &\tan \frac{\pi }{3}=\sqrt{3} \text{ (From}\frac{sin \theta}{cos \theta})\\ &\sec \frac{\pi }{3}=2 \text{ (From}\frac{1}{cos \theta})\\ &\csc \frac{\pi }{3}=\frac{2\sqrt{3}}{3}\text{ (From}\frac{1}{sin \theta} \text{ - do not forget to rationalize})\\ &\cot \frac{\pi }{3}=\frac{\sqrt{3}}{3}(\text{ (From}\frac{cos \theta}{sin \theta} \text{ - do not forget to rationalize})\end{align}$

Try It

Below are the values of all six trigonometric functions evaluated at common angle measures from Quadrant I.

Angle	$0$	$\frac{\pi }{6},\text{ or }{30}^{\circ}$	$\frac{\pi }{4},\text{ or } {45}^{\circ }$	$\frac{\pi }{3},\text{ or }{60}^{\circ }$	$\frac{\pi }{2},\text{ or }{90}^{\circ }$
Cosine	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
Sine	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1
Tangent	0	$\frac{\sqrt{3}}{3}$	1	$\sqrt{3}$	Undefined
Secant	1	$\frac{2\sqrt{3}}{3}$	$\sqrt{2}$	2	Undefined
Cosecant	Undefined	2	$\sqrt{2}$	$\frac{2\sqrt{3}}{3}$	1
Cotangent	Undefined	$\sqrt{3}$	1	$\frac{\sqrt{3}}{3}$	0

Limit Laws

The basic limit laws stated below, together with the other limit laws, allow us to evaluate limits of many algebraic functions.

Basic Limit Results

For any real number $a$ and any constant $c$ ,

$\underset{x\to a}{\lim}x=a$
$\underset{x\to a}{\lim}c=c$

Example: Evaluating a Basic Limit

Evaluate each of the following limits using the basic limit results above.

$\underset{x\to 2}{\lim}x$
$\underset{x\to 2}{\lim}5$

Show Solution

The limit of $x$ as $x$ approaches $a$ is $a$ : $\underset{x\to 2}{\lim}x=2$ .
The limit of a constant is that constant: $\underset{x\to 2}{\lim}5=5$ .

Try It

We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.

Limit Laws

Let $f(x)$ and $g(x)$ be defined for all $x\ne a$ over some open interval containing $a$ . Assume that $L$ and $M$ are real numbers such that $\underset{x\to a}{\lim}f(x)=L$ and $\underset{x\to a}{\lim}g(x)=M$ . Let $c$ be a constant. Then, each of the following statements holds:

Sum law for limits: $\underset{x\to a}{\lim}(f(x)+g(x))=\underset{x\to a}{\lim}f(x)+\underset{x\to a}{\lim}g(x)=L+M$

Difference law for limits: $\underset{x\to a}{\lim}(f(x)-g(x))=\underset{x\to a}{\lim}f(x)-\underset{x\to a}{\lim}g(x)=L-M$

Constant multiple law for limits: $\underset{x\to a}{\lim}cf(x)=c \cdot \underset{x\to a}{\lim}f(x)=cL$

Product law for limits: $\underset{x\to a}{\lim}(f(x) \cdot g(x))=\underset{x\to a}{\lim}f(x) \cdot \underset{x\to a}{\lim}g(x)=L \cdot M$

Quotient law for limits: $\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}=\dfrac{\underset{x\to a}{\lim}f(x)}{\underset{x\to a}{\lim}g(x)}=\frac{L}{M}$ for $M\ne 0$

Power law for limits: $\underset{x\to a}{\lim}(f(x))^n=(\underset{x\to a}{\lim}f(x))^n=L^n$ for every positive integer $n$ .

Root law for limits: $\underset{x\to a}{\lim}\sqrt[n]{f(x)}=\sqrt[n]{\underset{x\to a}{\lim}f(x)}=\sqrt[n]{L}$ for all $L$ if $n$ is odd and for $L\ge 0$ if $n$ is even

We now practice applying these limit laws to evaluate a limit.

Example: Evaluating a Limit Using Limit Laws

Use the limit laws to evaluate $\underset{x\to -3}{\lim}(4x+2)$ .

Show Solution

Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.

$\begin{array}{ccccc}\underset{x\to -3}{\lim}(4x+2)\hfill & =\underset{x\to -3}{\lim}4x+\underset{x\to -3}{\lim}2\hfill & & & \text{Apply the sum law.}\hfill \\ & =4 \cdot \underset{x\to -3}{\lim}x+\underset{x\to -3}{\lim}2\hfill & & & \text{Apply the constant multiple law.}\hfill \\ & =4 \cdot (-3)+2=-10\hfill & & & \text{Apply the basic limit results and simplify.}\hfill \end{array}$

Example: Using Limit Laws Repeatedly

Use the limit laws to evaluate $\underset{x\to 2}{\lim}\dfrac{2x^2-3x+1}{x^3+4}$ .

Show Solution

To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite the limit in terms of other limits, each new limit must exist for the limit law to be applied.

$\begin{array}{ccccc}\\ \\ \underset{x\to 2}{\lim}\large \frac{2x^2-3x+1}{x^3+4} & = \large \frac{\underset{x\to 2}{\lim}(2x^2-3x+1)}{\underset{x\to 2}{\lim}(x^3+4)} & & & \text{Apply the quotient law, making sure that} \, 2^3+4\ne 0 \\ & = \large \frac{2 \cdot \underset{x\to 2}{\lim}x^2-3 \cdot \underset{x\to 2}{\lim}x+\underset{x\to 2}{\lim}1}{\underset{x\to 2}{\lim}x^3+\underset{x\to 2}{\lim}4} & & & \text{Apply the sum law and constant multiple law.} \\ & = \large \frac{2 \cdot (\underset{x\to 2}{\lim}x)^2-3 \cdot \underset{x\to 2}{\lim}x+\underset{x\to 2}{\lim}1}{(\underset{x\to 2}{\lim}x)^3+\underset{x\to 2}{\lim}4} & & & \text{Apply the power law.} \\ & = \large \frac{2(4)-3(2)+1}{2^3+4}=\frac{1}{4} & & & \text{Apply the basic limit laws and simplify.} \end{array}$

Try It

Use the limit laws to evaluate $\underset{x\to 6}{\lim}(2x-1)\sqrt{x+4}$ . In each step, indicate the limit law applied.

Hint

Show Solution

$11\sqrt{10}$

Limits of Polynomial and Rational Functions

By now you have probably noticed that, in each of the previous examples, it has been the case that $\underset{x\to a}{\lim}f(x)=f(a)$ . This is not always true, but it does hold for all polynomials for any choice of $a$ and for all rational functions at all values of $a$ for which the rational function is defined.

Limits of Polynomial and Rational Functions

Let $p(x)$ and $q(x)$ be polynomial functions. Let $a$ be a real number. Then,

lim_{x \to a} p (x) = p (a)

$\underset{x\to a}{\lim}p(x)=p(a)$

lim_{x \to a} \frac{p (x)}{q (x)} = \frac{p (a)}{q (a)} when q (a) \neq 0

$\underset{x\to a}{\lim}\dfrac{p(x)}{q(x)}=\dfrac{p(a)}{q(a)} \, \text{when} \, q(a)\ne 0$

To see that this theorem holds, consider the polynomial $p(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0$ . By applying the sum, constant multiple, and power laws, we end up with

\begin{array}{cc} lim_{x \to a} p (x) & = lim_{x \to a} (c_{n} x^{n} + c_{n - 1} x^{n - 1} + \dots + c_{1} x + c_{0}) \\ = c_{n} (lim_{x \to a} x)^{n} + c_{n - 1} (lim_{x \to a} x)^{n - 1} + \dots + c_{1} (lim_{x \to a} x) + lim_{x \to a} c_{0} \\ = c_{n} a^{n} + c_{n - 1} a^{n - 1} + \dots + c_{1} a + c_{0} \\ = p (a) \end{array}

$\begin{array}{cc}\hfill \underset{x\to a}{\lim}p(x)& =\underset{x\to a}{\lim}(c_nx^n+c_{n-1}x^{n-1}+\cdots +c_1x+c_0)\hfill \\ & =c_n(\underset{x\to a}{\lim}x)^n+c_{n-1}(\underset{x\to a}{\lim}x)^{n-1}+\cdots +c_1(\underset{x\to a}{\lim}x)+\underset{x\to a}{\lim}c_0\hfill \\ & =c_na^n+c_{n-1}a^{n-1}+\cdots +c_1a+c_0\hfill \\ & =p(a)\hfill \end{array}$

It now follows from the quotient law that if $p(x)$ and $q(x)$ are polynomials for which $q(a)\ne 0$ , then

lim_{x \to a} \frac{p (x)}{q (x)} = \frac{p (a)}{q (a)}

$\underset{x\to a}{\lim}\dfrac{p(x)}{q(x)}=\dfrac{p(a)}{q(a)}$

The example below applies this result.

Example: Evaluating a Limit of a Rational Function

Evaluate the $\underset{x\to 3}{\lim}\dfrac{2x^2-3x+1}{5x+4}$ .

Show Solution

Since 3 is in the domain of the rational function $f(x)=\frac{2x^2-3x+1}{5x+4}$ , we can calculate the limit by substituting 3 for $x$ into the function. Thus,

lim_{x \to 3} \frac{2 x^{2} - 3 x + 1}{5 x + 4} = \frac{10}{19}

$\underset{x\to 3}{\lim}\dfrac{2x^2-3x+1}{5x+4}=\dfrac{10}{19}$

Try It

Evaluate $\underset{x\to -2}{\lim}(3x^3-2x+7)$ .

Hint

Show Solution

$−13$

Apply L’Hopital’s Rule

Indeterminate Form of Type $\frac{0}{0}$

L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider

lim_{x \to a} \frac{f (x)}{g (x)}

$\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}$

If $\underset{x\to a}{\lim}f(x)=L_1$ and $\underset{x\to a}{\lim}g(x)=L_2 \ne 0$ , then

lim_{x \to a} \frac{f (x)}{g (x)} = \frac{L_{1}}{L_{2}}

$\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}=\dfrac{L_1}{L_2}$

However, what happens if $\underset{x\to a}{\lim}f(x)=0$ and $\underset{x\to a}{\lim}g(x)=0$ ? We call this one of the indeterminate forms, of type $\frac{0}{0}$ . This is considered an indeterminate form because we cannot determine the exact behavior of $\frac{f(x)}{g(x)}$ as $x\to a$ without further analysis.

L’Hôpital’s Rule (0/0 Case)

Suppose $f$ and $g$ are differentiable functions over an open interval containing $a$ , except possibly at $a$ . If $\underset{x\to a}{\lim}f(x)=0$ and $\underset{x\to a}{\lim}g(x)=0$ , then

lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}

$\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}=\underset{x\to a}{\lim}\dfrac{f^{\prime}(x)}{g^{\prime}(x)}$ ,

assuming the limit on the right exists or is $\infty$ or $−\infty$ . This result also holds if we are considering one-sided limits, or if $a=\infty$ or $-\infty$ .

Example: Applying L’Hôpital’s Rule (0/0 Case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

$\underset{x\to 0}{\lim}\dfrac{1- \cos x}{x}$
$\underset{x\to 1}{\lim}\dfrac{\sin (\pi x)}{\ln x}$
$\underset{x\to \infty }{\lim}\dfrac{e^{\frac{1}{x}}-1}{\frac{1}{x}}$
$\underset{x\to 0}{\lim}\dfrac{\sin x-x}{x^2}$

Show Solution

Since the numerator $1- \cos x\to 0$ and the denominator $x\to 0$ , we can apply L’Hôpital’s rule to evaluate this limit. We have
$\begin{array}{ll} \underset{x\to 0}{\lim}\frac{1- \cos x}{x} & =\underset{x\to 0}{\lim}\frac{\frac{d}{dx}(1- \cos x)}{\frac{d}{dx}(x)} \\ & =\underset{x\to 0}{\lim}\frac{\sin x}{1} \\ & =\frac{\underset{x\to 0}{\lim}(\sin x)}{\underset{x\to 0}{\lim}(1)} \\ & =\frac{0}{1}=0 \end{array}$
As $x\to 1$ , the numerator $\sin (\pi x)\to 0$ and the denominator $\ln x \to 0$ . Therefore, we can apply L’Hôpital’s rule. We obtain
$\begin{array}{ll} \underset{x\to 1}{\lim}\frac{\sin (\pi x)}{\ln x} & =\underset{x\to 1}{\lim}\frac{\pi \cos (\pi x)}{1/x} \\ & =\underset{x\to 1}{\lim}(\pi x) \cos (\pi x) \\ & =(\pi \cdot 1)(-1)=−\pi \end{array}$
As $x\to \infty$ , the numerator $e^{1/x}-1\to 0$ and the denominator $(\frac{1}{x})\to 0$ . Therefore, we can apply L’Hôpital’s rule. We obtain
$\underset{x\to \infty }{\lim}\frac{e^{1/x}-1}{\frac{1}{x}}=\underset{x\to \infty }{\lim}\frac{e^{1/x}(\frac{-1}{x^2})}{(\frac{-1}{x^2})}=\underset{x\to \infty}{\lim} e^{1/x}=e^0=1$
As $x\to 0$ , both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s rule. We obtain
$\underset{x\to 0}{\lim}\frac{\sin x-x}{x^2}=\underset{x\to 0}{\lim}\frac{\cos x-1}{2x}$ .

Since the numerator and denominator of this new quotient both approach zero as $x\to 0$ , we apply L’Hôpital’s rule again. In doing so, we see that

$\underset{x\to 0}{\lim}\frac{\cos x-1}{2x}=\underset{x\to 0}{\lim}\frac{−\sin x}{2}=0$ .

Therefore, we conclude that

$\underset{x\to 0}{\lim}\frac{\sin x-x}{x^2}=0$ .

Try It

Evaluate $\underset{x\to 0}{\lim}\dfrac{x}{\tan x}$ .

Hint

$\frac{d}{dx} \tan x= \sec ^2 x$

Show Solution

$1$

Indeterminate Form of Type $\frac{\infty}{\infty}$

We can also use L’Hôpital’s rule to evaluate limits of quotients $\frac{f(x)}{g(x)}$ in which $f(x)\to \pm \infty$ and $g(x)\to \pm \infty$ . Limits of this form are classified as indeterminate forms of type $\infty / \infty$ . Again, note that we are not actually dividing $\infty$ by $\infty$ . Since $\infty$ is not a real number, that is impossible; rather, $\infty / \infty$ is used to represent a quotient of limits, each of which is $\infty$ or $−\infty$ .

L’Hôpital’s Rule ( $\infty / \infty$ Case)

Suppose $f$ and $g$ are differentiable functions over an open interval containing $a$ , except possibly at $a$ . Suppose $\underset{x\to a}{\lim}f(x)=\infty$ (or $−\infty$ ) and $\underset{x\to a}{\lim}g(x)=\infty$ (or $−\infty$ ). Then,

lim_{x \to a} \frac{f (x)}{g (x)} = lim_{x \to a} \frac{f^{'} (x)}{g^{'} (x)}

$\underset{x\to a}{\lim}\dfrac{f(x)}{g(x)}=\underset{x\to a}{\lim}\dfrac{f^{\prime}(x)}{g^{\prime}(x)}$ ,

assuming the limit on the right exists or is $\infty$ or $−\infty$ . This result also holds if the limit is infinite, if $a=\infty$ or $−\infty$ , or the limit is one-sided.

Example: Applying L’Hôpital’s Rule ( $\infty /\infty$ Case)

Evaluate each of the following limits by applying L’Hôpital’s rule.

$\underset{x\to \infty }{\lim}\dfrac{3x+5}{2x+1}$
$\underset{x\to 0^+}{\lim}\dfrac{\ln x}{\cot x}$

Show Solution

Since $3x+5$ and $2x+1$ are first-degree polynomials with positive leading coefficients, $\underset{x\to \infty }{\lim}(3x+5)=\infty$ and $\underset{x\to \infty }{\lim}(2x+1)=\infty$ . Therefore, we apply L’Hôpital’s rule and obtain
$\underset{x\to \infty}{\lim}\dfrac{3x+5}{2x+\frac{1}{x}}=\underset{x\to \infty }{\lim}\dfrac{3}{2}=\dfrac{3}{2}$ .

Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the module we showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of $x$ in the denominator. In doing so, we saw that

$\underset{x\to \infty }{\lim}\dfrac{3x+5}{2x+1}=\underset{x\to \infty }{\lim}\dfrac{3+\frac{5}{x}}{2+\frac{1}{x}}=\dfrac{3}{2}$ .

L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
Here, $\underset{x\to 0^+}{\lim} \ln x=−\infty$ and $\underset{x\to 0^+}{\lim} \cot x=\infty$ . Therefore, we can apply L’Hôpital’s rule and obtain
$\underset{x\to 0^+}{\lim}\dfrac{\ln x}{\cot x}=\underset{x\to 0^+}{\lim}\dfrac{\frac{1}{x}}{−\csc^2 x}=\underset{x\to 0^+}{\lim}\dfrac{1}{−x \csc^2 x}$ .

Now as $x\to 0^+$ , $\csc^2 x\to \infty$ . Therefore, the first term in the denominator is approaching zero and the second term is getting really large. In such a case, anything can happen with the product. Therefore, we cannot make any conclusion yet. To evaluate the limit, we use the definition of $\csc x$ to write

$\underset{x\to 0^+}{\lim}\dfrac{1}{−x \csc^2 x}=\underset{x\to 0^+}{\lim}\dfrac{\sin^2 x}{−x}$ .

Now $\underset{x\to 0^+}{\lim} \sin^2 x=0$ and $\underset{x\to 0^+}{\lim} x=0$ , so we apply L’Hôpital’s rule again. We find

$\underset{x\to 0^+}{\lim}\dfrac{\sin^2 x}{−x}=\underset{x\to 0^+}{\lim}\dfrac{2 \sin x \cos x}{-1}=\dfrac{0}{-1}=0$ .

We conclude that

$\underset{x\to 0^+}{\lim}\dfrac{\ln x}{\cot x}=0$ .

Try It

Evaluate $\underset{x\to \infty }{\lim}\dfrac{\ln x}{5x}$

Hint

$\frac{d}{dx}\ln x=\frac{1}{x}$

Show Solution

L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms $\frac{0}{0}$ and $\frac{\infty}{\infty}$ . However, we can also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The expressions $0 \cdot \infty$ , $\infty - \infty$ , $1^{\infty}$ , $\infty^0$ , and $0^0$ are all considered indeterminate forms. These expressions are not real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite the indeterminate forms in such a way that we arrive at the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .

Indeterminate Form of Type $0 \cdot \infty$

Suppose we want to evaluate $\underset{x\to a}{\lim}(f(x) \cdot g(x))$ , where $f(x)\to 0$ and $g(x)\to \infty$ (or $−\infty$ ) as $x\to a$ . Since one term in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to the product. We use the notation $0 \cdot \infty$ to denote the form that arises in this situation. The expression $0 \cdot \infty$ is considered indeterminate because we cannot determine without further analysis the exact behavior of the product $f(x)g(x)$ as $x\to {a}$ . For example, let $n$ be a positive integer and consider

f (x) = \frac{1}{(x^{n} + 1)}

$f(x)=\dfrac{1}{(x^n+1)}$ and

g (x) = 3 x^{2}

$g(x)=3x^2$ .

As $x\to \infty$ , $f(x)\to 0$ and $g(x)\to \infty$ . However, the limit as $x\to \infty$ of $f(x)g(x)=\frac{3x^2}{(x^n+1)}$ varies, depending on $n$ . If $n=2$ , then $\underset{x\to \infty }{\lim}f(x)g(x)=3$ . If $n=1$ , then $\underset{x\to \infty }{\lim}f(x)g(x)=\infty$ . If $n=3$ , then $\underset{x\to \infty }{\lim}f(x)g(x)=0$ . Here we consider another limit involving the indeterminate form $0 \cdot \infty$ and show how to rewrite the function as a quotient to use L’Hôpital’s rule.

Example: Indeterminate Form of Type $0·\infty$

Evaluate $\underset{x\to 0^+}{\lim}x \ln x$

Show Solution

First, rewrite the function $x \ln x$ as a quotient to apply L’Hôpital’s rule. If we write

x \ln x = \frac{\ln x}{1 / x}

$x \ln x=\frac{\ln x}{1/x}$ ,

we see that $\ln x\to −\infty$ as $x\to 0^+$ and $\frac{1}{x}\to \infty$ as $x\to 0^+$ . Therefore, we can apply L’Hôpital’s rule and obtain

lim_{x \to 0^{+}} \frac{\ln x}{1 / x} = lim_{x \to 0^{+}} \frac{\frac{d}{d x} (\ln x)}{\frac{d}{d x} (1 / x)} = lim_{x \to 0^{+}} \frac{1 / x}{- 1 / x^{2}} = lim_{x \to 0^{+}} (- x) = 0

$\underset{x\to 0^+}{\lim}\frac{\ln x}{1/x}=\underset{x\to 0^+}{\lim}\frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(1/x)}=\underset{x\to 0^+}{\lim}\frac{1/x}{-1/x^2}=\underset{x\to 0^+}{\lim}(−x)=0$ .

We conclude that

lim_{x \to 0^{+}} x \ln x = 0

$\underset{x\to 0^+}{\lim}x \ln x=0$ .

The function y = x ln(x) is graphed for values x ≥ 0. At x = 0, the value of the function is 0.

Finding the limit at $x=0$ of the function $f(x)=x \ln x$ .

Try It

Evaluate $\underset{x\to 0}{\lim}x \cot x$

Hint

Write $x \cot x=\frac{x \cos x}{\sin x}$

Show Solution

Indeterminate Form of Type $\infty -\infty$

Another type of indeterminate form is $\infty -\infty$ . Consider the following example. Let $n$ be a positive integer and let $f(x)=3x^n$ and $g(x)=3x^2+5$ . As $x\to \infty$ , $f(x)\to \infty$ and $g(x)\to \infty$ . We are interested in $\underset{x\to \infty}{\lim}(f(x)-g(x))$ . Depending on whether $f(x)$ grows faster, $g(x)$ grows faster, or they grow at the same rate, as we see next, anything can happen in this limit. Since $f(x)\to \infty$ and $g(x)\to \infty$ , we write $\infty -\infty$ to denote the form of this limit. As with our other indeterminate forms, $\infty -\infty$ has no meaning on its own and we must do more analysis to determine the value of the limit. For example, suppose the exponent $n$ in the function $f(x)=3x^n$ is $n=3$ , then

lim_{x \to \infty} (f (x) - g (x)) = lim_{x \to \infty} (3 x^{3} - 3 x^{2} - 5) = \infty

$\underset{x\to \infty }{\lim}(f(x)-g(x))=\underset{x\to \infty }{\lim}(3x^3-3x^2-5)=\infty$ .

On the other hand, if $n=2$ , then

lim_{x \to \infty} (f (x) - g (x)) = lim_{x \to \infty} (3 x^{2} - 3 x^{2} - 5) = - 5

$\underset{x\to \infty }{\lim}(f(x)-g(x))=\underset{x\to \infty }{\lim}(3x^2-3x^2-5)=-5$ .

However, if $n=1$ , then

lim_{x \to \infty} (f (x) - g (x)) = lim_{x \to \infty} (3 x - 3 x^{2} - 5) = - \infty

$\underset{x\to \infty }{\lim}(f(x)-g(x))=\underset{x\to \infty }{\lim}(3x-3x^2-5)=−\infty$ .

Therefore, the limit cannot be determined by considering only $\infty -\infty$ . Next we see how to rewrite an expression involving the indeterminate form $\infty -\infty$ as a fraction to apply L’Hôpital’s rule.

Example: Indeterminate Form of Type $\infty -\infty$

Evaluate $\underset{x\to 0^+}{\lim}\left(\dfrac{1}{x^2}-\dfrac{1}{\tan x}\right)$ .

Show Solution

By combining the fractions, we can write the function as a quotient. Since the least common denominator is $x^2 \tan x$ , we have

\frac{1}{x^{2}} - \frac{1}{\tan x} = \frac{(\tan x) - x^{2}}{x^{2} \tan x}

$\frac{1}{x^2}-\frac{1}{\tan x}=\frac{(\tan x)-x^2}{x^2 \tan x}$

As $x\to 0^+$ , the numerator $\tan x-x^2 \to 0$ and the denominator $x^2 \tan x \to 0$ . Therefore, we can apply L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have

lim_{x \to 0^{+}} \frac{(\tan x) - x^{2}}{x^{2} \tan x} = lim_{x \to 0^{+}} \frac{(\sec^{2} x) - 2 x}{x^{2} \sec^{2} x + 2 x \tan x}

$\underset{x\to 0^+}{\lim}\frac{(\tan x)-x^2}{x^2 \tan x}=\underset{x\to 0^+}{\lim}\frac{(\sec^2 x)-2x}{x^2 \sec^2 x+2x \tan x}$

As $x\to 0^+$ , $(\sec^2 x)-2x \to 1$ and $x^2 \sec^2 x+2x \tan x \to 0$ . Since the denominator is positive as $x$ approaches zero from the right, we conclude that

lim_{x \to 0^{+}} \frac{(\sec^{2} x) - 2 x}{x^{2} \sec^{2} x + 2 x \tan x} = \infty

$\underset{x\to 0^+}{\lim}\frac{(\sec^2 x)-2x}{x^2 \sec^2 x+2x \tan x}=\infty$

Therefore,

lim_{x \to 0^{+}} (\frac{1}{x^{2}} - \frac{1}{t a n x}) = \infty

$\underset{x\to 0^+}{\lim}(\frac{1}{x^2}-\frac{1}{ tan x})=\infty$

Try It

Evaluate $\underset{x\to 0^+}{\lim}\left(\dfrac{1}{x}-\dfrac{1}{\sin x}\right)$ .

Hint

Show Solution

Other Types of Indeterminate Form

Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions $0^0$ , $\infty^0$ , and $1^{\infty}$ are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise when finding limits. Now we examine how L’Hôpital’s rule can be used to evaluate limits involving these indeterminate forms.

Since L’Hôpital’s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want to evaluate $\underset{x\to a}{\lim}f(x)^{g(x)}$ and we arrive at the indeterminate form $\infty^0$ . (The indeterminate forms $0^0$ and $1^{\infty}$ can be handled similarly.)

Example: Indeterminate Form of Type $\infty^0$

Evaluate $\underset{x\to \infty }{\lim} x^{\frac{1}{x}}$

Show Solution

Let $y=x^{1/x}$ . Then,

\ln (x^{1 / x}) = \frac{1}{x} \ln x = \frac{\ln x}{x}

$\ln (x^{1/x})=\frac{1}{x} \ln x=\frac{\ln x}{x}$

We need to evaluate $\underset{x\to \infty }{\lim}\frac{\ln x}{x}$ . Applying L’Hôpital’s rule, we obtain

lim_{x \to \infty} \ln y = lim_{x \to \infty} \frac{\ln x}{x} = lim_{x \to \infty} \frac{1 / x}{1} = 0

$\underset{x\to \infty }{\lim} \ln y=\underset{x\to \infty}{\lim}\frac{\ln x}{x}=\underset{x\to \infty}{\lim}\frac{1/x}{1}=0$

Therefore, $\underset{x\to \infty }{\lim}\ln y=0$ . Since the natural logarithm function is continuous, we conclude that

\ln (lim_{x \to \infty} y) = 0

$\ln (\underset{x\to \infty}{\lim} y)=0$ ,

which leads to

lim_{x \to \infty} y = lim_{x \to \infty} \frac{\ln x}{x} = e^{0} = 1

$\underset{x\to \infty }{\lim} y=\underset{x\to \infty}{\lim}\frac{\ln x}{x}=e^0=1$

Hence,

lim_{x \to \infty} x^{1 / x} = 1

$\underset{x\to \infty}{\lim} x^{1/x}=1$

Try It

Evaluate $\underset{x\to \infty}{\lim} x^{\frac{1}{\ln x}}$

Hint

Let $y=x^{1/ \ln x}$ and apply the natural logarithm to both sides of the equation.

Show Solution

$e$

Example: Indeterminate Form of Type $0^0$

Evaluate $\underset{x\to 0^+}{\lim} x^{\sin x}$

Show Solution

Let

y = x^{\sin x}

$y=x^{\sin x}$

Therefore,

\ln y = \ln (x^{\sin x}) = \sin x \ln x

$\ln y=\ln (x^{\sin x})= \sin x \ln x$

We now evaluate $\underset{x\to 0^+}{\lim} \sin x \ln x$ . Since $\underset{x\to 0^+}{\lim} \sin x=0$ and $\underset{x\to 0^+}{\lim} \ln x=−\infty$ , we have the indeterminate form $0 \cdot \infty$ . To apply L’Hôpital’s rule, we need to rewrite $\sin x \ln x$ as a fraction. We could write

\sin x \ln x = \frac{\sin x}{1 / \ln x}

$\sin x \ln x=\frac{\sin x}{1/ \ln x}$

\sin x \ln x = \frac{\ln x}{1 / \sin x} = \frac{\ln x}{\csc x}

$\sin x \ln x=\frac{\ln x}{1/ \sin x}=\frac{\ln x}{\csc x}$

Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain

lim_{x \to 0^{+}} \sin x \ln x = lim_{x \to 0^{+}} \frac{\sin x}{1 / \ln x} = lim_{x \to 0^{+}} \frac{\cos x}{- 1 / (x (\ln x)^{2})} = lim_{x \to 0^{+}} (- x (\ln x)^{2} \cos x)

$\underset{x\to 0^+}{\lim} \sin x \ln x=\underset{x\to 0^+}{\lim}\frac{\sin x}{1/ \ln x}=\underset{x\to 0^+}{\lim}\frac{\cos x}{-1/(x(\ln x)^2)}=\underset{x\to 0^+}{\lim}(−x(\ln x)^2 \cos x)$

Unfortunately, we not only have another expression involving the indeterminate form $0 \cdot \infty$ , but the new limit is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By writing

\sin x \ln x = \frac{\ln x}{1 / \sin x} = \frac{\ln x}{\csc x}

$\sin x \ln x=\frac{\ln x}{1/ \sin x}=\frac{\ln x}{\csc x}$

and applying L’Hôpital’s rule, we obtain

lim_{x \to 0^{+}} \sin x \ln x = lim_{x \to 0^{+}} \frac{\ln x}{\csc x} = lim_{x \to 0^{+}} \frac{1 / x}{- \csc x \cot x} = lim_{x \to 0^{+}} \frac{- 1}{x \csc x \cot x}

$\underset{x\to 0^+}{\lim} \sin x \ln x=\underset{x\to 0^+}{\lim}\frac{\ln x}{\csc x}=\underset{x\to 0^+}{\lim}\frac{1/x}{− \csc x \cot x}=\underset{x\to 0^+}{\lim}\frac{-1}{x \csc x \cot x}$

Using the fact that $\csc x=\frac{1}{\sin x}$ and $\cot x=\frac{\cos x}{\sin x}$ , we can rewrite the expression on the right-hand side as

lim_{x \to 0^{+}} \frac{- \sin^{2} x}{x \cos x} = lim_{x \to 0^{+}} [\frac{\sin x}{x} \cdot (- \tan x)] = (lim_{x \to 0^{+}} \frac{\sin x}{x}) \cdot (lim_{x \to 0^{+}} (- \tan x)) = 1 \cdot 0 = 0

$\underset{x\to 0^+}{\lim}\frac{−\sin^2 x}{x \cos x}=\underset{x\to 0^+}{\lim}[\frac{\sin x}{x} \cdot (−\tan x)]=(\underset{x\to 0^+}{\lim}\frac{\sin x}{x}) \cdot (\underset{x\to 0^+}{\lim}(−\tan x))=1 \cdot 0=0$

We conclude that $\underset{x\to 0^+}{\lim} \ln y=0$ . Therefore, $\ln (\underset{x\to 0^+}{\lim} y)=0$ and we have

lim_{x \to 0^{+}} y = lim_{x \to 0^{+}} x^{\sin x} = e^{0} = 1

$\underset{x\to 0^+}{\lim} y=\underset{x\to 0^+}{\lim} x^{\sin x}=e^0=1$

Hence,

lim_{x \to 0^{+}} x^{\sin x} = 1

$\underset{x\to 0^+}{\lim} x^{\sin x}=1$

Try It

Evaluate $\underset{x\to 0^+}{\lim} x^x$

Hint

Let $y=x^x$ and take the natural logarithm of both sides of the equation.

Show Solution

Vector-Valued Functions: Prerequisite Material