I have to find the following limits:
a) $\displaystyle\lim_{x\to \infty}x\left(\left(1+\frac {1}{x}\right)^x-e\right)$.
b)$\displaystyle\lim_{x\to 0^+}\left(\frac {\sin x}{x}\right)^{\frac {1}{x^2}}$
I had thought to solve this using l'Hôpital's Rule.
But for a I obtained that of recurring type. So I could not proceed through that.
As I had put $t=1/x$, so $\frac {(1+t)^{1/t}-e}{t}$ as $t \to 0$
By applying the Rule I got $\frac {-(1+t)^{1/t}ln(1+t)}{t^2}$. Again applying the Rule I obtained a recurring cycle.
Is there another way to do this or did I do any mistake?
b) For b I obtained an answer using the series of $\sin x$ and using the formula of the exponential. That leads to my answer as $e^{-1/6}$. Is it possible to do this example using L'Hôpital's Rule efficiently?
The idea of doing $x=1/t$ is good: $$ \lim_{t\to0^+}\frac{(1+t)^{1/t}-e}{t} $$ If $f(t)=(1+t)^{1/t}$, then, for $t>0$, $$ f'(t)=(1+t)^{1/t}\frac{t-(1+t)\log(1+t)}{t^2(1+t)} $$ Since $\displaystyle\lim_{t\to0^+}\frac{(1+t)^{1/t}}{1+t}=e$, you are reduced to computing $$ \lim_{t\to0^+}\frac{t-(1+t)\log(1+t)}{t^2}= \lim_{t\to0^+}\frac{1-1-\log(1+t)}{2t}=-\frac{1}{2} $$
For the second limit, compute instead the limit of the logarithm, that is, $$ \lim_{x\to0^+}\frac{\log\dfrac{\sin x}{x}}{x^2} $$ Since $\frac{\sin x}{x}=1-\frac{x^2}{6}+o(x^2)$, you get $$ \lim_{x\to0^+}\frac{-x^2/6+o(x^2)}{x^2}=-\frac{1}{6} $$ So your given limit is $e^{-1/6}$.
You can also do it with l'Hôpital, but it's not as pretty: $$ \lim_{x\to0^+}\frac{\dfrac{\cos x}{\sin x}-\dfrac{1}{x}}{2x}= \lim_{x\to0^+}\frac{x\cos x-\sin x}{2x^2\sin x}= \lim_{x\to0^+}\frac{-x\sin x}{4x\sin x+2x^2\cos x}= -\frac{1}{2}\lim_{x\to0^+}\frac{\sin x}{2\sin x+x\cos x}= -\frac{1}{2}\lim_{x\to0^+}\frac{\cos x}{3\cos x-x\sin x} $$