Math Snap

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 $(1/x) - 7 \rightarrow -7$

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. $\lim _{x \rightarrow \infty} \frac{\tan ^{-1}(x)}{(1 / x)-7} = -\frac{\pi}{14}$