Solve $\lim_{x\to 0} (\ln (x+e))^{\cot x}$ without l'Hôpital $$\lim_{x\to 0}  (\ln (x+e))^{\cot x}$$ 
Since we have an indeterminate form of $1^{\infty }$, we should simplify in a way that we can transform the expression into e to the power of something, but I can't find a way. 
The use of l'Hôpital's rule or series is prohibited.
 A: Rewrite it as
$$(\ln(x+e))^{\cot(x)}=\exp\left[x\cot(x)\frac{\ln(\ln(x+e))}x\right]$$
And use the known limit
$$\lim_{x\to0}\frac{\sin(x)}x=1$$
To get
$$\lim_{x\to0}x\cot(x)=\lim_{x\to0}\cos(x)\left(\frac{\sin(x)}x\right)^{-1}=1\cdot1^{-1}=1$$
and then use the definition of the derivative to compute
$$\lim_{x\to0}\frac{\ln(\ln(x+e))}x=\frac d{dx}\ln(\ln(x+e))\bigg|_{x=0}$$
Evaluating the derivative with chain rule and
$$\frac d{dx}\ln(x)=\frac1x$$
which should give you a final result of
$$\lim_{x\to0}(\ln(x+e))^{\cot(x)}=e^{1/e}$$
A: Taking the natural log you have 
$$\cot x \ln \ln(x+e))=$$
$$\cos x \frac{x}{\sin x} \frac{\ln \ln (x+e)}{\ln (x+e)-1}\frac{\ln(x+e)-1}{x}$$
The initial factors all limit to $1$, as for the last,
$$\frac{\ln(x+e)-1}{x}=
\frac{\ln(x+e)-\ln e}{x}=\frac{\ln(\frac{x}{e}+1)}{x}$$
$$=\frac{1}{e}\frac{\ln(\frac{x}{e}+1)}{x/e}\to\frac{1}{e}$$
A: Note that:
$$\ln(x+e)=\ln e +\ln\left(1+\frac{x}{e}\right)=1+\frac{x}{e}+o(x)$$
Thus:

$$(\ln (x+e))^{\cot x}=\left[\left(1+\frac{x}{e}+o(x)\right)^\frac1x\right]^\frac{x}{tanx}\to e^{\frac1e}$$

A: First of all, you want to compute the limit of the logarithm of your function:
$$
\lim_{x\to0}\cot x\ln(\ln(x+e))
$$
The first step is almost obvious: note that
$$
\lim_{x\to0}x\cot x=\lim_{x\to0}\frac{x}{\sin x}\cos x=1
$$
so you can reduce to computing
$$
\lim_{x\to0}\frac{\ln(\ln(x+e))}{x}
$$
because if this limit exists it will be equal to the one you want to compute.
How do we do this one? Let's do the substitution $x=et$, so it becomes
$$
\lim_{t\to0}\frac{\ln(\ln e+\ln(1+t))}{et}=
\lim_{t\to0}\frac{\ln(1+\ln(1+t))}{et}
$$
Not yet in the best form. Let's do a new substitution: $\ln(1+t)=u$, that is, $t=e^u-1$, so we get
$$
\frac{1}{e}\lim_{u\to0}\frac{\ln(1+u)}{e^u-1}=
\frac{1}{e}\lim_{u\to0}\frac{\ln(1+u)}{u}\frac{u}{e^u-1}=…
$$

 Since $$\lim_{u\to0}\frac{\ln(1+u)}{u}=1\qquad\lim_{u\to0}\frac{u}{e^u-1}=1$$ are both known limits, you can conclude that $$\lim_{x\to0}\cot x\ln(\ln(x+e))=\frac{1}{e}$$ Therefore $$\lim_{x\to0}(\ln(x+e))^{\cot x}=e^{1/e}$$

Of course, using l'Hôpital is easier and doesn't require imagination:
$$
\lim_{x\to0}\frac{\ln(\ln(x+e))}{\tan x}=
\lim_{x\to0}\frac{\dfrac{1}{\ln(x+e)}\dfrac{1}{x+e}}{1+\tan^2x}=\frac{1}{e}
$$
A: You have
\begin{align}
\ln(e+x)^{\cot x}&=\ln\left(e(1+\frac xe)\right)^{\frac{\cos x}{\sin x}}=\left(1+\ln(1+\frac xe)\right)^{\frac{\cos x}{\sin x}}\\ \ \\
&=\left(1+\frac xe+o(x^2) \right)^{\frac{\cos x}{\sin x}}\\ \ \\
&=\exp\left(\frac{\cos x}{\sin x}\,\ln\left(1+\frac xe+o(x^2) \right) \right)\\ \ \\
&=\exp\left(\frac{\cos x}{\sin x}\,\left(\frac xe+o(x^2)\right)\right)\\ \ \\
&=\exp\left(\frac{1+o(x^2)}{x+o(x^3)}\,\left(\frac xe+o(x^2)\right)\right)\\ \ \\
&=\exp\left(\frac{1+o(x^2)}{1+o(x^2)}\,\left(\frac 1e+o(x)\right)\right)\\ \ \\
&\xrightarrow[x\to0]{}\exp\left(\frac1e\right)=e^{1/e}.
\end{align}
The relevant Taylor polynomials used above are 
$$
\ln(1+x)=x+o(x^2),\ \ \cos x=1+o(x^2),\ \ \ \sin x=x+o(x^3),
$$
together with $a^b=e^{b\ln a}$ (which is the definition of $a^b$)
