Finding a limit using L'Hospital's rule Find the limit
$$\lim_{x\to\infty} (e^x+x)^{4/x}.$$
Use L'Hospital's rule if appropriate. If there is a more elementary method, consider using it.
 A: Hint: For a more elementary method, I suggest Squeezing. Note that for positive $x$ we have
$$e^x\lt e^x+x\lt 2e^x$$
A: $$\lim_{x\to\infty} (e^x +x)^{4/x}=\exp{\lim_{x\to\infty} \frac{4\ln(e^x +x)}{x}}$$
A: You could take logs, e.g. denoting your limit by $\ell$ we have
$$\log\ell = \lim_{x\to\infty}\frac{4\log(e^x+x)}{x}.$$
Using L'Hopital's rule this gives
$$\log\ell = \lim_{x\to\infty}\frac{4 \left(e^x+1\right)}{x+e^x}=\lim_{x\to\infty}\frac{4 \left(e^x+x\right)}{x+e^x}+\frac{4 \left(1-x\right)}{x+e^x}.$$ But this is just $$4+\lim_{x\to\infty}\frac{4 \left(1-x\right)}{x+e^x}=4.$$
Hence, $$\ell = e^4.$$
A: Let:
$$y = \lim_{x\to\infty} (e^x+x)^{4/x}$$
Then,
$$\ln y = \lim_{x\to\infty} \frac{4}{x}\ln(e^x+x)$$
Direct substitution yields $\frac{\infty}{\infty}$. Using L'Hosptial's Rule:
$$\ln y = \lim_{x\to\infty} \frac{4}{x}\ln(e^x+x) = \frac{\frac{4(e^x+1)}{e^x+x}}{1} = \frac{4(e^x+1)}{e^x+x} = \frac{4e^x}{e^x+1} = \frac{4e^x}{e^x} = 4$$
So, $\ln y = 4$, and $y = e^4$
Note that multiple dervatives were taken since the direct substitution yielded $\frac{\infty}{\infty}$
