Finding $\lim_{n \to \infty} n (e^{1/n} - 1)$ 
Find $\lim_{n \to \infty} n (e^{1/n} - 1)$.

Could anyone give me a hint please?
 A: If we let $x=1/n$ we get
$$\lim_{x\to0}\frac{e^x-1}{x}.$$
A: MVT:
$\displaystyle {\int_{0}^{1/n}}e^x dx =e^{1/n} -1 =(1/n) e^t,$  $t \in [0,1/n].$
$1\le n(e^{1/n}-1) =e^t \le e^{1/n}.$
A: $$\lim_{n\to\infty} n(e^{\frac1n}-1)=\lim_{n\to\infty}\int_0^1 e^{\frac{x}{n}}\ dx=\int_0^1 \lim_{n\to\infty} e^{\frac{x}{n}}\ dx=\int_0^1\ dx=1$$
A: Without any reference to derivative concept, recall that
$$ \left(1+\frac1n\right)^n\le e\le  \left(1+\frac1n\right)^{n+1}$$
and by squeeze theorem we have 
$$ 1=\frac{\left[\left(1+\frac1n\right)^n\right]^{1/n} -1}{\frac1n}\le \frac{e^{1/n} -1}{\frac1n}\le  \frac{\left[\left(1+\frac1n\right)^{n+1}\right]^{1/n} -1}{\frac1n} \to 1$$
indeed by Bernoulli's inequality $(1+x)^r \le 1+rx$ for $0\le r\le 1$ we have
$$\frac{\left[\left(1+\frac1n\right)^{n+1}\right]^{1/n} -1}{\frac1n}= \frac{\left(1+\frac1n\right)\left(1+\frac1n\right)^{1/n} -1}{\frac1n}\le \frac{\left(1+\frac1n\right)\left(1+\frac1{n^2}\right) -1}{\frac1n}=$$
$$=\frac{\frac1{n}+\frac1{n^2}+\frac1{n^3}}{\frac1n}=1+\frac1{n}+\frac1{n^2}\to 1$$
