question 7 - 1 point\nevaluate the following limit using lhospitals rule.\n\\(\\lim_{x\\to\\infty}(5x)e^{-2x}…

Explanation:

Step1: Rewrite the limit

We have $\lim_{x\rightarrow\infty}(5x)e^{-2x}=\lim_{x\rightarrow\infty}\frac{5x}{e^{2x}}$. As $x\rightarrow\infty$, we get the indeterminate - form $\frac{\infty}{\infty}$, so we can apply L'Hospital's rule.

Step2: Apply L'Hospital's rule

Differentiate the numerator and denominator. The derivative of $5x$ with respect to $x$ is $5$, and the derivative of $e^{2x}$ with respect to $x$ is $2e^{2x}$. So, $\lim_{x\rightarrow\infty}\frac{5x}{e^{2x}}=\lim_{x\rightarrow\infty}\frac{5}{2e^{2x}}$.

Step3: Evaluate the new limit

As $x\rightarrow\infty$, $e^{2x}\rightarrow\infty$. Then $\lim_{x\rightarrow\infty}\frac{5}{2e^{2x}} = 0$.

Answer:

$0$