Question

4.6 limits at infinity and horizontal asymp - l2: problem 4 (6 points) results for this submission 1 of the answers is not correct. evaluate the following limits. (a) $lim_{x
ightarrowinfty}\frac{4}{e^{x}+7}=0$ (b) $lim_{x
ightarrow-infty}\frac{4}{e^{x}+7}=$ note: if needed, enter inf or infinity for $infty$ and -inf or -infinity for $-infty$. note: in order to get credit for this problem all answers must be correct. note: you are in the reduced scoring period. all work counts for 85% of the original. preview my answers submit answers your score was recorded. scores are sent to d2l brightspace every 24 hours. you have attempted this problem 3 times. you received a score of 0% for this attempt. your overall recorded score is 0%. you have 2 attempts remaining.

Explanation:

Step1: Analyze limit as $x\to-\infty$

As $x\to-\infty$, $e^{x}\to0$. Because the exponential - function $y = e^{x}$ has the property that when $x$ approaches negative infinity, the value of $e^{x}$ gets closer and closer to 0.

Step2: Calculate the limit

We have $\lim_{x\to-\infty}\frac{4}{e^{x}+7}$. Substitute the limit of $e^{x}$ as $x\to-\infty$ into the expression. Since $\lim_{x\to-\infty}e^{x}=0$, then $\lim_{x\to-\infty}\frac{4}{e^{x}+7}=\frac{4}{0 + 7}=\frac{4}{7}$.

Answer:

$\frac{4}{7}$