Find these limits using l'Hopital's rule I could use some help solving these 2 problems. I've finished the rest, I just need help on these two. I know I'm supposed to use natural log and l'hopitals rule. 


*

*As the $x$ goes to $0$, find the limit of $\tan(x)^{\sin(x)}$.

*As the $x$ goes to infinity, find the limit of $(e^x+x)^{1/x}$
I tried to use natural log for both of these and got no where. What am I missing? I appreciate any input. It's very much needed. 
 A: HINT: note that $$\tan(x)^{\sin(x)}=e^{\sin(x)\ln(\tan(x))}$$
and $$(e^x+x)^{1/x}=e^\frac{\ln(e^x+x)}{x}$$
A: For the first, take $f(x) = (\tan x)^{\sin x}$. Then 
$$\log f(x) = \sin x\log\tan x=\sin x(\log \sin x -\log \cos x)$$
As $x\rightarrow 0$, $-\sin x \cdot \log \cos x\rightarrow0$. Using L'Hôpital we check that
$$\lim_{x\rightarrow0}\log f(x) = \lim_{x\rightarrow0}\frac{\log \sin x}{\frac1{\sin x}}=\lim_{x\rightarrow0}\frac{\frac{\cos x}{\sin x}}{-\frac{\cos x}{(\sin x)^2}}=\lim_{x\rightarrow0}-\sin x = 0$$
Thus, $\lim_{x\rightarrow0} f(x) = e^0=1$
For the second, $g(x) = (x+e^x)^{1/x}$
$$\log g(x) = \frac1x(x+e^x)=1+\frac{e^x}x$$
Then
$$\lim_{x\rightarrow0} \log g(x) = 1+\lim_{x\rightarrow0}\frac{e^x}x=1+1=2$$
Thus $\lim_{x\rightarrow0} g(x) = e^2$
A: Sorry for answering on an older question, but it seemed that there was an unnoticed error in one of the answers. For $$\log  g(x) = \frac1x(x+e^x)=1+\frac{e^x}x$$
the $log$ on $(x+e^x)$ disappears, meaning we cannot use the distributive property here. Instead, $$\log g(x) = \frac1x\log(x+e^x)$$
Therefore, $$\lim_{x \to \infty} \log g(x) = \lim_{x \to \infty}\frac1x\log(x+e^x) = \lim_{x \to \infty} \frac{e^x +1}{e^x+x}= \lim_{x \to \infty} \frac{e^x}{e^x+1}= \lim_{x \to \infty} \frac{e^x}{e^x}= \lim_{x \to \infty} 1 = 1$$
(Here we use L'Hôpital's rule thrice, each justified by the indeterminate form $\infty / \infty$.) Finally, $$\lim_{x \to \infty} g(x) = e^1 = e$$
A: It's maybe worth noting that the second limit can be done without L'Hopital, provided you know, for example, that
$$1\le\left(1+{1\over u}\right)^u\le 3$$
for all (large) $u$.  (The inequality holds with $e$ in place of $3$, but there are elementary inductive proofs for the $3$.  Any upper bound would serve our purpose here.)  We have
$$(e^x+x)^{1/x}=e\left(1+{x\over e^x}\right)^{1/x}=e\left[\left(1+{x\over e^x}\right)^{e^x/x}\right]^{1/e^x}$$
and thus
$$e\le(e^x+x)^{1/x}\le e\cdot3^{1/e^x}$$
Since $3^{1/e^x}\to3^0=1$ as $x\to\infty$, the squeeze theorem tells us
$$\lim_{x\to\infty}(e^x+x)^{1/x}=e$$
