Find $\lim _{x \rightarrow 0}\left[x /\left(e^{x}-1\right)\right]^{1 / x}$
It seems obvious that the initial step should be taken using an exponential function, where the new form yields $\frac{\ln(\frac{x}{e^x-1})}{x}$. However, L'Hospital's rule does not work in this case. Are there any possible alternatives to transform the given expression?
Best Answer
In fact, $\dfrac{x}{e^{x}-1}\rightarrow 1$ by L'Hopital.
So $\log\dfrac{x}{e^{x}-1}\rightarrow\log 1=0$, so the limit in question is in L'Hopital form.
It becomes \begin{align*} &\lim\dfrac{1}{\dfrac{x}{e^{x}-1}}\dfrac{(e^{x}-1)-xe^{x}}{(e^{x}-1)^{2}}\\ &=\lim\dfrac{e^{x}-1-xe^{x}}{x(e^{x}-1)}\\ &=\lim\dfrac{e^{x}-e^{x}-xe^{x}}{e^{x}-1+xe^{x}}\\ &=\lim\dfrac{-xe^{x}}{e^{x}-1+xe^{x}}\\ &=\lim\dfrac{-e^{x}-xe^{x}}{e^{x}+e^{x}+xe^{x}}\\ &=\lim\dfrac{-1-x}{2+x}\\ &=-\dfrac{1}{2}. \end{align*}