14. -/1 points details my notes sesscalcet2 3.7.038. find the limit. use lhospitals rule if appropriate. if…

Explanation:

Step1: Take the natural - log of the function

Let $y=(e^{x}+x)^{\frac{2}{x}}$. Then $\ln y=\frac{2}{x}\ln(e^{x}+x)=\frac{2\ln(e^{x}+x)}{x}$.

Step2: Find the limit of $\ln y$ as $x\rightarrow\infty$

As $x\rightarrow\infty$, we have the indeterminate form $\frac{\infty}{\infty}$. So we can apply L'Hopital's Rule. Differentiate the numerator and denominator: The derivative of the numerator $u = 2\ln(e^{x}+x)$ using the chain - rule is $u^\prime=\frac{2(e^{x}+1)}{e^{x}+x}$. The derivative of the denominator $v = x$ is $v^\prime = 1$. So $\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}\frac{2(e^{x}+1)}{e^{x}+x}$.

Step3: Apply L'Hopital's Rule again (if needed)

As $x\rightarrow\infty$, we still have the indeterminate form $\frac{\infty}{\infty}$. Differentiate the numerator and denominator of $\frac{2(e^{x}+1)}{e^{x}+x}$ again. The derivative of the numerator $u_1=2(e^{x}+1)$ is $u_1^\prime = 2e^{x}$. The derivative of the denominator $v_1=e^{x}+x$ is $v_1^\prime=e^{x}+1$. So $\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}\frac{2e^{x}}{e^{x}+1}$.

Step4: Simplify the limit of $\ln y$

Divide both the numerator and denominator by $e^{x}$: $\lim_{x\rightarrow\infty}\frac{2e^{x}}{e^{x}+1}=\lim_{x\rightarrow\infty}\frac{2}{1 + \frac{1}{e^{x}}}=2$.

Step5: Find the limit of $y$

Since $\lim_{x\rightarrow\infty}\ln y = 2$, and $y = e^{\ln y}$, then $\lim_{x\rightarrow\infty}y=e^{2}$.

Answer:

$e^{2}$