Evaluate $\lim_{x\to(e^{-1})^+}\frac{e^{\frac{\ln(1+\ln x)}x}}{x-e^{-1}}$ 
Evaluate $\lim\limits_{x\to(e^{-1})^+}\dfrac{e^{\dfrac{\ln(1+\ln x)}x}}{x-e^{-1}}$

It's a $\dfrac00$ form.
Approach$1$: L'Hopital$$\lim_{x\to(e^{-1})^+}\dfrac{e^{\dfrac{\ln(1+\ln x)}x}\cdot\dfrac{\dfrac1{1+\ln x}-\ln(1+\ln x)}{x^2}}1\\=\lim_{x\to(e^{-1})^+}e^{\dfrac{\ln(1+\ln x)}x}\cdot\dfrac{1-\ln(1+\ln x)^{1+\ln x}}{x^2(1+\ln x)}$$
Again this is indeterminate. So, I abandoned this approach.
Approach$2$: Series expansion.$$\ln(1+\ln x)=\ln x-\frac{(\ln x)^2}2+\cdots$$
This is to be divided by $x$. Don't think I am going anywhere with this.
Any hint please?
 A: First let $t=x-\frac 1e$. Then we have-
$$L=\lim_{t \to 0^{+}} \frac {\left( \ln(1+et)\right)^{\frac {e}{et+1}}}{t}$$ since for any possible $k$, $e^{\ln k}=k$.
Now let $et=u$, then limit turns into:
$$L=e\lim_{u \to 0^{+}} \frac {\left(\ln(1+u)\right)^{\frac {e}{1+u}}}{u}=e\lim_{u \to 0^{+}} \left(\frac {\ln(1+u)}{u}\right)^{\frac {e}{1+u}} u^{\frac {e}{1+u}-1}$$
Now from the well known result that $\frac {\ln(1+x)}{x}=1-\frac x2 +O(x^2)$, we can substitute value of $u$ to evaluate limit and find that $$L=0$$

Perhaps it is note-worthy here that your approach 2 is, in fact, incorrect. Taylor series for $\ln(1+x)$ should only be written around $x=0$, and not in any neighborhood of $x$. However in my solution you'll notice that the series has been applied keeping in mind the conditions.
A: Let $1+\log x =t$ so that $t\to 0^+$. The expression under limit is then transformed into $$e\cdot\frac{e^{(\log t) /e^{t-1}} }{e^t-1} =e\cdot\frac{t^{e^{1-t}}} {t}\cdot\frac{e^t-1}{t}$$ The last fraction tends to $1$ and hence desired limit is equal to the limit of $$et^{e^{1-t}-1}$$ The exponent in above expression tends to $e-1>0$ and the base $t\to 0^+$ hence $t^{e^{1-t}-1}\to 0$. The desired limit is thus $0$.
