Taking the limit of n(e^1/n −1) as n approaches infinity then proving it by the squeeze theorem

Question

Instead of using L'Hospital's rule can this be proved by using the definition of the limit and or the squeeze theorem.

Answer 1

Use Mean Value Theorem to deduce that $1+u+u^{2}/2\leq e^{u}\leq 1+u+u^{2}$ for $0\leq u\leq 1/2$, then for $n\geq 2$, we have $1/n\leq 2$ and hence $1+(1/n)+1/(2n^{2})\leq e^{1/n}\leq 1+(1/n)+1/n^{2}$, so $n(e^{1/n}-1)\rightarrow 1$.

The use of Mean Value Theorem here:

Let $\varphi(u)=1+u+u^{2}-e^{u}$, $0\leq u\leq 1/2$, then for $0<u\leq 1/2$, $\varphi(u)=\varphi(u)-\varphi(0)=\varphi'(\xi)u$, where $\xi$ lies in between $0$ and $u$.

Since $\varphi'(\xi)=1+2\xi-e^{\xi}$, we investigate the function $\eta(u)=1+2u-e^{u}$ for $0\leq u\leq 1/2$. Note that for $0<u\leq 1/2$, $\eta(u)=\eta(u)-\eta(0)=\eta'(\omega)u=(2-e^{\omega})u>0$ for some $\omega$ lies in between $0$ and $u$, in particular, $\omega$ lies in between $0$ and $1/2$.

So $\eta(\xi)>0$ and hence $\varphi'(\xi)=\eta(\xi)>0$, so $\varphi(u)>0$.

Answer 2

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$1+x\le e^x\le \frac1{1-x}\tag 1$$

for $x<1$.

Letting $x=\frac1n$ for $n>1$ in $(1)$, subtracting $1$, and multiplying the result by $n$ reveals

$$1\le n(e^{1/n}-1)\le \frac{1}{1-\frac1n}$$

whence applying the squeeze theorem yields the coveted result.