Math / Calculus

QuestionCompute the following limits using L'Hospital's rule if appropriate. Use INF to denote oo and MINF to denote $\begin{array}{l} \lim _{x \rightarrow 1} \frac{7^{\infty}-7}{x^{2}-1}=\square \\ \lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}= \end{array}$ $\square$
Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

1. We are given two limit problems to solve.
2. L'Hospital's Rule can be applied when the limit results in an indeterminate form such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
3. We need to check if L'Hospital's Rule is applicable for each limit.

STEP 2

1. Evaluate the first limit $\lim _{x \rightarrow 1} \frac{7^{\infty}-7}{x^{2}-1}$ .
2. Evaluate the second limit $\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}$ .

STEP 3

Evaluate the expression $7^{\infty} - 7$ :
Since $7^{\infty}$ is a constant (infinity), the expression $7^{\infty} - 7$ simplifies to $\infty$ .

STEP 4

Evaluate the denominator as $x \rightarrow 1$ :
$x^2 - 1 \rightarrow 1^2 - 1 = 0$

STEP 5

Determine if L'Hospital's Rule is applicable:
The limit $\lim _{x \rightarrow 1} \frac{7^{\infty}-7}{x^{2}-1}$ results in $\frac{\infty}{0}$ , which is not an indeterminate form that allows the use of L'Hospital's Rule.
Therefore, the limit is $\infty$ .

STEP 6

Evaluate the limit $\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}$ :
First, evaluate the numerator and denominator separately as $x \rightarrow \infty$ .

STEP 7

Evaluate the numerator $\tan^{-1}(x)$ as $x \rightarrow \infty$ :
$\tan^{-1}(x) \rightarrow \frac{\pi}{2}$

STEP 8

Evaluate the denominator $(1/x) - 7$ as $x \rightarrow \infty$ :
$(1/x) \rightarrow 0$ , so $(1/x) - 7 \rightarrow -7$

STEP 9

Determine if L'Hospital's Rule is applicable:
The limit $\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}$ results in $\frac{\frac{\pi}{2}}{-7}$ , which is not an indeterminate form.
Therefore, the limit is $-\frac{\pi}{14}$ .
The solutions to the limits are:
1. $\lim _{x \rightarrow 1} \frac{7^{\infty}-7}{x^{2}-1} = \text{INF}$
2. $\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7} = -\frac{\pi}{14}$

Was this helpful?

Studdy Solution

STEP 1

1. We are given two limit problems to solve.
2. L'Hospital's Rule can be applied when the limit results in an indeterminate form such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
3. We need to check if L'Hospital's Rule is applicable for each limit.

STEP 2

1. Evaluate the first limit $\lim _{x \rightarrow 1} \frac{7^{\infty}-7}{x^{2}-1}$ .
2. Evaluate the second limit $\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}$ .

STEP 3

Evaluate the expression $7^{\infty} - 7$ :
Since $7^{\infty}$ is a constant (infinity), the expression $7^{\infty} - 7$ simplifies to $\infty$ .

STEP 4

Evaluate the denominator as $x \rightarrow 1$ :
$x^2 - 1 \rightarrow 1^2 - 1 = 0$

STEP 5

Determine if L'Hospital's Rule is applicable:
The limit $\lim _{x \rightarrow 1} \frac{7^{\infty}-7}{x^{2}-1}$ results in $\frac{\infty}{0}$ , which is not an indeterminate form that allows the use of L'Hospital's Rule.
Therefore, the limit is $\infty$ .

STEP 6

Evaluate the limit $\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}$ :
First, evaluate the numerator and denominator separately as $x \rightarrow \infty$ .

STEP 7

Evaluate the numerator $\tan^{-1}(x)$ as $x \rightarrow \infty$ :
$\tan^{-1}(x) \rightarrow \frac{\pi}{2}$

STEP 8

Evaluate the denominator $(1/x) - 7$ as $x \rightarrow \infty$ :
$(1/x) \rightarrow 0$ , so $(1/x) - 7 \rightarrow -7$

STEP 9

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

1. We are given two limit problems to solve.2. L'Hospital's Rule can be applied when the limit results in an indeterminate form such as 00 \frac{0}{0} 00​ or ∞∞ \frac{\infty}{\infty} ∞∞​.3. We need to check if L'Hospital's Rule is applicable for each limit.

STEP 2

1. Evaluate the first limit lim⁡x→17∞−7x2−1\lim _{x \rightarrow 1} \frac{7^{\infty}-7}{x^{2}-1}limx→1​x2−17∞−7​.2. Evaluate the second limit lim⁡x→∞tan⁡−1(x)(1/x)−7\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}limx→∞​(1/x)−7tan−1(x)​.

STEP 3

Evaluate the expression 7∞−77^{\infty} - 77∞−7:Since 7∞7^{\infty}7∞ is a constant (infinity), the expression 7∞−77^{\infty} - 77∞−7 simplifies to ∞\infty∞.

STEP 4

Evaluate the denominator as x→1x \rightarrow 1x→1:x2−1→12−1=0 x^2 - 1 \rightarrow 1^2 - 1 = 0 x2−1→12−1=0

STEP 5

STEP 6

Evaluate the limit lim⁡x→∞tan⁡−1(x)(1/x)−7\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}limx→∞​(1/x)−7tan−1(x)​:First, evaluate the numerator and denominator separately as x→∞x \rightarrow \inftyx→∞.

STEP 7

Evaluate the numerator tan⁡−1(x)\tan^{-1}(x)tan−1(x) as x→∞x \rightarrow \inftyx→∞:tan⁡−1(x)→π2\tan^{-1}(x) \rightarrow \frac{\pi}{2}tan−1(x)→2π​

STEP 8

Evaluate the denominator (1/x)−7(1/x) - 7(1/x)−7 as x→∞x \rightarrow \inftyx→∞:(1/x)→0(1/x) \rightarrow 0(1/x)→0, so (1/x)−7→−7(1/x) - 7 \rightarrow -7(1/x)−7→−7

STEP 9

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. We are given two limit problems to solve.
2. L'Hospital's Rule can be applied when the limit results in an indeterminate form such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$ .
3. We need to check if L'Hospital's Rule is applicable for each limit.

1. Evaluate the first limit $\lim _{x \rightarrow 1} \frac{7^{\infty}-7}{x^{2}-1}$ .
2. Evaluate the second limit $\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}$ .

Evaluate the expression $7^{\infty} - 7$ :
Since $7^{\infty}$ is a constant (infinity), the expression $7^{\infty} - 7$ simplifies to $\infty$ .

Evaluate the denominator as $x \rightarrow 1$ :
$x^2 - 1 \rightarrow 1^2 - 1 = 0$

Evaluate the limit $\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7}$ :
First, evaluate the numerator and denominator separately as $x \rightarrow \infty$ .

Evaluate the numerator $\tan^{-1}(x)$ as $x \rightarrow \infty$ :
$\tan^{-1}(x) \rightarrow \frac{\pi}{2}$

Evaluate the denominator $(1/x) - 7$ as $x \rightarrow \infty$ :
$(1/x) \rightarrow 0$ , so $(1/x) - 7 \rightarrow -7$