How to find $\lim_{x \to \infty} x(e^{1/x}-1)^x$ without L'Hopital's rule or Taylor expansion?

Question

I'm trying to find the following limit $$ \lim_{x \to \infty} x(e^{1/x}-1)^x,$$ without L'Hopital or Taylor.

So far I got

\begin{align*} \lim_{x \to \infty} x(e^{1/x}-1)^x &= \lim_{x \to \infty} x(e^{1/x}-1)^x \\ &= \lim_{u \to 0} \frac{1}{\ln(u+1)}(u)^{\frac{1}{\ln(u+1)}} \\ &= \lim_{u \to 0} \frac{1}{\ln(u+1)} \left((u)^{\frac{u}{ \ln(u+1)}}\right)^{\frac{1}{u} } \\ &= \lim_{u \to 0} \frac{1}{\ln(u+1)} \left((u)^{\frac{1}{ \ln(u+1)^{1/u}}}\right)^{\frac{1}{u} } \\ \end{align*}

Any hint for going on?

Answer 1

Take logs: $\ln x + x \ln (e^{1/x}-1) \to-\infty$. So limit is $0$.

Answer 2

By standard limit $t^t\to 1$ as $t\to 0$ we have

$$x(e^{1/x}-1)^x=\frac{(e^{1/x}-1)^x}{\frac1x}=\left(\frac{e^{1/x}-1}{\left(\frac1{x}\right)^\frac1x}\right)^x \to \left[\left(\frac{0}{1}\right)^\infty=\right]0$$

Answer 3

Your approach is promising and can be easily completed. Just note that $u/\log(1+u)\to 1$ so one can replace first factor $1/\log(1+u)$ with $1/u$ to get $$\lim_{u\to 0^{+}}u^{1/\log(1+u)-1}$$ The exponent tends to $\infty$ and hence the desired limit is $0$. In case you want to be more formal (as some instructors may insist on it) just note that if $0<u<1/3$ we have $\log(1+u)\leq u$ so that $$\frac{1}{\log(1+u)}-1\geq \frac{1}{u}-1\geq 2$$ and therefore expression under limit lies between $0$ and $u^2$ if $0<u<1/3$. Proof is complete now by squeeze theorem.