Skills Review for Sequences

Learning Outcomes

Write the terms of a sequence defined by a recursive formula
Calculate the limit of a function as 𝑥 increases or decreases without bound
Recognize when to apply L’Hôpital’s rule

In the Sequences section, we will look at ordered lists of numbers (sequences) and determine whether they converge or diverge. Here we will review how to evaluate a recursive (recurrence) formula, take limits at infinity, and apply L’Hôpital’s Rule.

Use a Recursive Formula

A recursive formula is a formula that defines its value at a particular input using the result of the previous input(s).

A recursive formula always has two parts: the value of an initial input and an equation defining each term in terms of preceding terms. For example, suppose we know the following:

\begin{matrix} a_{1} = 3 a_{n} = 2 a_{n - 1} - 1, for n \geq 2 \end{matrix}

$\begin{align}&{a}_{1}=3 \\ &{a}_{n}=2{a}_{n - 1}-1, \text{ for } n\ge 2 \end{align}$

We can find the subsequent terms of the recursive formula using the first term.

\begin{matrix} a_{1} = 3 a_{2} = 2 a_{1} - 1 = 2 (3) - 1 = 5 a_{3} = 2 a_{2} - 1 = 2 (5) - 1 = 9 a_{4} = 2 a_{3} - 1 = 2 (9) - 1 = 17 \end{matrix}

$\begin{align}&{a}_{1}=3\\ &{a}_{2}=2{a}_{1}-1=2\left(3\right)-1=5\\ &{a}_{3}=2{a}_{2}-1=2\left(5\right)-1=9\\ &{a}_{4}=2{a}_{3}-1=2\left(9\right)-1=17\end{align}$

So, the first four terms are $3,5,9,\text{ and},17$ .

How To: Given a recursive formula with only the first term provided, write the first $n$ terms of a sequence.

Identify the initial term, ${a}_{1}$ , which is given as part of the formula. This is the first term.
To find the second term, ${a}_{2}$ , substitute the initial term into the formula for ${a}_{n - 1}$ . Solve.
To find the third term, ${a}_{3}$ , substitute the second term into the formula. Solve.
Repeat until you have solved for the $n\text{th}$ term.

Example: Writing the Terms of a Sequence Defined by a Recursive Formula

Write the first five terms of the sequence defined by the recursive formula.

$\begin{align} {a}_{1}&=9 \\ {a}_{n}&=3{a}_{n - 1}-20\text{, for }n\ge 2 \end{align}$

Show Solution

The first term is given in the formula. For each subsequent term, we replace ${a}_{n - 1}$ with the value of the preceding term.

$\begin{align}&n=1 && {a}_{1}=9 \\ &n=2 && {a}_{2}=3{a}_{1}-20=3\left(9\right)-20=27 - 20=7 \\ &n=3 && {a}_{3}=3{a}_{2}-20=3\left(7\right)-20=21 - 20=1 \\ &n=4 && {a}_{4}=3{a}_{3}-20=3\left(1\right)-20=3 - 20=-17 \\ &n=5 && {a}_{5}=3{a}_{4}-20=3\left(-17\right)-20=-51 - 20=-71 \end{align}$

The first five terms are $\left\{9,7,1,-17,-71\right\}$

Try It

Write the first five terms of the sequence defined by the recursive formula.

$\begin{align}{a}_{1}&=2\\ {a}_{n}&=2{a}_{n - 1}+1\text{, for }n\ge 2\end{align}$

Show Solution

$\left\{2, 5, 11, 23, 47\right\}$

Try It

Take Limits at Infinity

Recall that $\underset{x \to a}{\lim}f(x)=L$ means $f(x)$ becomes arbitrarily close to $L$ as long as $x$ is sufficiently close to $a$ . We can extend this idea to limits at infinity. For example, consider the function $f(x)=2+\frac{1}{x}$ . As can be seen graphically in Figure 1 and numerically in the table beneath it, as the values of $x$ get larger, the values of $f(x)$ approach 2. We say the limit as $x$ approaches $\infty$ of $f(x)$ is 2 and write $\underset{x\to \infty }{\lim}f(x)=2$ . Similarly, for $x<0$ , as the values $|x|$ get larger, the values of $f(x)$ approaches 2. We say the limit as $x$ approaches $−\infty$ of $f(x)$ is 2 and write $\underset{x\to a}{\lim}f(x)=2$ .

The function f(x) 2 + 1/x is graphed. The function starts negative near y = 2 but then decreases to −∞ near x = 0. The function then decreases from ∞ near x = 0 and gets nearer to y = 2 as x increases. There is a horizontal line denoting the asymptote y = 2.

Figure 1. The function approaches the asymptote $y=2$ as $x$ approaches $\pm \infty$ .

Values of a function $f$ as $x \to \pm \infty$
$x$	10	100	1,000	10,000
$2+\frac{1}{x}$	2.1	2.01	2.001	2.0001
$x$	-10	-100	-1000	-10,000
$2+\frac{1}{x}$	1.9	1.99	1.999	1.9999

More generally, for any function $f$ , we say the limit as $x \to \infty$ of $f(x)$ is $L$ if $f(x)$ becomes arbitrarily close to $L$ as long as $x$ is sufficiently large. In that case, we write $\underset{x\to \infty}{\lim}f(x)=L$ . Similarly, we say the limit as $x\to −\infty$ of $f(x)$ is $L$ if $f(x)$ becomes arbitrarily close to $L$ as long as $x<0$ and $|x|$ is sufficiently large. In that case, we write $\underset{x\to −\infty }{\lim}f(x)=L$ . We now look at the definition of a function having a limit at infinity.

Definition

(Informal) If the values of $f(x)$ become arbitrarily close to $L$ as $x$ becomes sufficiently large, we say the function $f$ has a limit at infinity and write

lim x \to \infty f (x) = L

$\underset{x\to \infty }{\lim}f(x)=L$

If the values of $f(x)$ becomes arbitrarily close to $L$ for $x<0$ as $|x|$ becomes sufficiently large, we say that the function $f$ has a limit at negative infinity and write

lim x \to - \infty f (x) = L

$\underset{x\to -\infty }{\lim}f(x)=L$

If the values $f(x)$ are getting arbitrarily close to some finite value $L$ as $x\to \infty$ or $x\to −\infty$ , the graph of $f$ approaches the line $y=L$ . In that case, the line $y=L$ is a horizontal asymptote of $f$ (Figure 2). For example, for the function $f(x)=\frac{1}{x}$ , since $\underset{x\to \infty }{\lim}f(x)=0$ , the line $y=0$ is a horizontal asymptote of $f(x)=\frac{1}{x}$ .

Definition

If $\underset{x\to \infty }{\lim}f(x)=L$ or $\underset{x \to −\infty}{\lim}f(x)=L$ , we say the line $y=L$ is a horizontal asymptote of $f$ .

The figure is broken up into two figures labeled a and b. Figure a shows a function f(x) approaching but never touching a horizontal dashed line labeled L from above. Figure b shows a function f(x) approaching but never a horizontal dashed line labeled M from below.

Figure 2. (a) As $x\to \infty$ , the values of $f$ are getting arbitrarily close to $L$ . The line $y=L$ is a horizontal asymptote of $f$ . (b) As $x\to −\infty$ , the values of $f$ are getting arbitrarily close to $M$ . The line $y=M$ is a horizontal asymptote of $f$ .

A function cannot cross a vertical asymptote because the graph must approach infinity (or negative infinity) from at least one direction as $x$ approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may cross a horizontal asymptote an unlimited number of times. For example, the function $f(x)=\frac{ \cos x}{x}+1$ shown in Figure 3 intersects the horizontal asymptote $y=1$ an infinite number of times as it oscillates around the asymptote with ever-decreasing amplitude.

The function f(x) = (cos x)/x + 1 is shown. It decreases from (0, ∞) and then proceeds to oscillate around y = 1 with decreasing amplitude.

Figure 3. The graph of $f(x)=\cos x/x+1$ crosses its horizontal asymptote $y=1$ an infinite number of times.

Example: Computing Limits at Infinity

For each of the following functions $f$ , evaluate $\underset{x\to \infty }{\lim}f(x)$ and $\underset{x\to −\infty }{\lim}f(x)$ .

$f(x)=5-\frac{2}{x^2}$
$f(x)=\dfrac{\sin x}{x}$
$f(x)= \tan^{-1} (x)$

Show Solution

Using the algebraic limit laws, we have $\underset{x\to \infty }{\lim}(5-\frac{2}{x^2})=\underset{x\to \infty }{\lim}5-2(\underset{x\to \infty }{\lim}\frac{1}{x})(\underset{x\to \infty }{\lim}\frac{1}{x})=5-2 \cdot 0=5$ . Similarly, $\underset{x\to -\infty }{\lim}f(x)=5$ .
Since $-1\le \sin x\le 1$ for all $x$ , we have
$\frac{-1}{x}\le \frac{\sin x}{x}\le \frac{1}{x}$

for all $x \ne 0$ . Also, since

$\underset{x\to \infty }{\lim}\frac{-1}{x}=0=\underset{x\to \infty }{\lim}\frac{1}{x}$ ,

we can apply the squeeze theorem to conclude that

$\underset{x\to \infty }{\lim}\frac{\sin x}{x}=0$

Similarly,

$\underset{x\to −\infty}{\lim}\frac{\sin x}{x}=0$
To evaluate $\underset{x\to \infty }{\lim} \tan^{-1} (x)$ and $\underset{x\to −\infty}{\lim} \tan^{-1} (x)$ , we first consider the graph of $y= \tan (x)$ over the interval $(−\pi /2,\pi /2)$ as shown in the following graph.

Figure 6. The graph of $\tan x$ has vertical asymptotes at $x=\pm \frac{\pi }{2}$

Since

lim x \to (π / 2)^{-} tan x = \infty

$\underset{x\to (\pi/2)^-}{\lim} \tan x=\infty$ ,

it follows that

lim x \to \infty {tan}^{- 1} (x) = \frac{π}{2}

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

Similarly, since

lim x \to (π / 2)^{+} tan x = - \infty

$\underset{x\to (\pi/2)^+}{\lim} \tan x=−\infty$ ,

it follows that

lim x \to - \infty {tan}^{- 1} (x) = - \frac{π}{2}

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

Try It

Evaluate $\underset{x\to −\infty}{\lim}\left(3+\frac{4}{x}\right)$ and $\underset{x\to \infty }{\lim}\left(3+\frac{4}{x}\right)$ . Determine the horizontal asymptotes of $f(x)=3+\frac{4}{x}$ , if any.

Hint

$\underset{x\to \pm \infty }{\lim}1/x=0$

Show Solution

Infinite Limits at Infinity

Sometimes the values of a function $f$ become arbitrarily large as $x\to \infty$ (or as $x\to −\infty )$ . In this case, we write $\underset{x\to \infty }{\lim}f(x)=\infty$ (or $\underset{x\to −\infty }{\lim}f(x)=\infty )$ . On the other hand, if the values of $f$ are negative but become arbitrarily large in magnitude as $x\to \infty$ (or as $x\to −\infty )$ , we write $\underset{x\to \infty }{\lim}f(x)=−\infty$ (or $\underset{x\to −\infty }{\lim}f(x)=−\infty )$ .

For example, consider the function $f(x)=x^3$ . As seen in the table below and Figure 2, as $x\to \infty$ the values $f(x)$ become arbitrarily large. Therefore, $\underset{x\to \infty }{\lim}x^3=\infty$ . On the other hand, as $x\to −\infty$ , the values of $f(x)=x^3$ are negative but become arbitrarily large in magnitude. Consequently, $\underset{x\to −\infty }{\lim}x^3=−\infty$ .

Values of a power function as $x\to \pm \infty$
$x$	10	20	50	100	1000
$x^3$	1000	8000	125,000	1,000,000	1,000,000,000
$x$	-10	-20	-50	-100	-1000
$x^3$	-1000	-8000	-125,000	-1,000,000	-1,000,000,000

The function f(x) = x3 is graphed. It is apparent that this function rapidly approaches infinity as x approaches infinity.

Figure 2. For this function, the functional values approach infinity as $x\to \pm \infty$ .

Definition

(Informal) We say a function $f$ has an infinite limit at infinity and write

lim x \to \infty f (x) = \infty

$\underset{x\to \infty }{\lim}f(x)=\infty$

if $f(x)$ becomes arbitrarily large for $x$ sufficiently large. We say a function has a negative infinite limit at infinity and write

lim x \to \infty f (x) = - \infty

$\underset{x\to \infty }{\lim}f(x)=−\infty$

if $f(x)<0$ and $|f(x)|$ becomes arbitrarily large for $x$ sufficiently large. Similarly, we can define infinite limits as $x\to −\infty$ .

Try It

Find $\underset{x\to \infty }{\lim}3x^2$ .

Show Solution

$\infty$

Try It

Apply L’Hôpital’s Rule

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. We have seen examples of this earlier in the text. For example, consider

lim x \to 2 \frac{x^{2} - 4}{x - 2}

$\underset{x\to 2}{\lim}\dfrac{x^2-4}{x-2}$ and

lim x \to 0 \frac{sin x}{x}

$\underset{x\to 0}{\lim}\dfrac{ \sin x}{x}$

For the first of these examples, we can evaluate the limit by factoring the numerator and writing

lim x \to 2 \frac{x^{2} - 4}{x - 2} = lim x \to 2 \frac{(x + 2) (x - 2)}{x - 2} = lim x \to 2 (x + 2) = 2 + 2 = 4

$\underset{x\to 2}{\lim}\dfrac{x^2-4}{x-2}=\underset{x\to 2}{\lim}\dfrac{(x+2)(x-2)}{x-2}=\underset{x\to 2}{\lim}(x+2)=2+2=4$

For $\underset{x\to 0}{\lim}\frac{\sin x}{x}$ we were able to show, using a geometric argument, that

lim x \to 0 \frac{sin x}{x} = 1

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

Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could not calculate previously.

The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions $f$ and $g$ such that $\underset{x\to a}{\lim}f(x)=0=\underset{x\to a}{\lim}g(x)$ and such that $g^{\prime}(a)\ne 0$ For $x$ near $a$ , we can write

f (x) \approx f (a) + f^{'} (a) (x - a)

$f(x)\approx f(a)+f^{\prime}(a)(x-a)$

and

g (x) \approx g (a) + g^{'} (a) (x - a)

$g(x)\approx g(a)+g^{\prime}(a)(x-a)$

Therefore,

\frac{f (x)}{g (x)} \approx \frac{f (a) + f^{'} (a) (x - a)}{g (a) + g^{'} (a) (x - a)}

$\dfrac{f(x)}{g(x)}\approx \dfrac{f(a)+f^{\prime}(a)(x-a)}{g(a)+g^{\prime}(a)(x-a)}$

Two functions y = f(x) and y = g(x) are drawn such that they cross at a point above x = a. The linear approximations of these two functions y = f(a) + f’(a)(x – a) and y = g(a) + g’(a)(x – a) are also drawn.

Figure 1. If $\underset{x\to a}{\lim}f(x)=\underset{x\to a}{\lim}g(x)$ , then the ratio $f(x)/g(x)$ is approximately equal to the ratio of their linear approximations near $a$ .

Since $f$ is differentiable at $a$ , then $f$ is continuous at $a$ , and therefore $f(a)=\underset{x\to a}{\lim}f(x)=0$ . Similarly, $g(a)=\underset{x\to a}{\lim}g(x)=0$ . If we also assume that $f^{\prime}$ and $g^{\prime}$ are continuous at $x=a$ , then $f^{\prime}(a)=\underset{x\to a}{\lim}f^{\prime}(x)$ and $g^{\prime}(a)=\underset{x\to a}{\lim}g^{\prime}(x)$ . Using these ideas, we conclude that

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

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

Note that the assumption that $f^{\prime}$ and $g^{\prime}$ are continuous at $a$ and $g^{\prime}(a)\ne 0$ can be loosened. We state L’Hôpital’s rule formally for the indeterminate form $\frac{0}{0}$ . Also note that the notation $\frac{0}{0}$ does not mean we are actually dividing zero by zero. Rather, we are using the notation $\frac{0}{0}$ to represent a quotient of limits, each of which is zero.

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

$1$

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

Sequences and Series: Prerequisite Material