this question is designed to be answered without a calculator. which statement best describes lim xe^x? x→…

this question is designed to be answered without a calculator. which statement best describes lim xe^x? x→ -∞ the limit exists, and its value is 0. the limit exists, and its value is 1. the limit does not exist, as lhopitals rule does not apply to products. the limit does not exist, as the application of lhopitals rule repeatedly yields the form 0/0.

this question is designed to be answered without a calculator. which statement best describes lim xe^x? x→ -∞ the limit exists, and its value is 0. the limit exists, and its value is 1. the limit does not exist, as lhopitals rule does not apply to products. the limit does not exist, as the application of lhopitals rule repeatedly yields the form 0/0.

Answer

Answer:

The limit exists, and its value is 0.

Explanation:

Step1: Rewrite the limit

We have $\lim_{x\rightarrow-\infty}xe^{x}=\lim_{x\rightarrow-\infty}\frac{x}{e^{-x}}$.

Step2: Apply L'Hopital's rule

As $x\rightarrow-\infty$, we have the indeterminate form $\frac{-\infty}{\infty}$. By L'Hopital's rule, $\lim_{x\rightarrow-\infty}\frac{x}{e^{-x}}=\lim_{x\rightarrow-\infty}\frac{1}{-e^{-x}}$.

Step3: Evaluate the new - limit

As $x\rightarrow-\infty$, $e^{-x}\rightarrow\infty$. So, $\lim_{x\rightarrow-\infty}\frac{1}{-e^{-x}} = 0$.