What is the limit $ \displaystyle \lim_{x\to\infty}\bigg[x^2(\mathrm{e}^{\frac{1}{x}}-\mathrm{e}^{\frac{1}{x+1}})\bigg] $?

Question

How do I solve this limit? $$\lim_{x\to\infty}\left[x^2(\mathrm{e}^{\frac{1}{x}}-\mathrm{e}^{\frac{1}{x+1}})\right] $$

Can you try to do it without Taylor expansion or l'Hopital would be wonderful.

Thanks for trying.

Answer 1

$x^2(e^{\frac{1}{x}}-e^{\frac{1}{x+1}})=x^2e^{\frac{1}{x}}(1-e^{\frac{-1}{x(x+1)}})\sim x^2e^{\frac{1}{x}}\frac{1}{x(x+1)}\sim e^{\frac{1}{x}}\to_{x \to \infty}1$

Answer 2

$$x^2\int_{\frac{1}{x+1}}^{\frac{1}{x}}e^t\,dt = \frac{x^2}{x(x+1)}+O\left(x^2 \int_{\frac{1}{x+1}}^{\frac{1}{x}}t\,dt\right) = \color{red}{\large 1}+O\left(\frac{1}{x}\right) $$ as $x\to +\infty$.

Answer 3

I don't know for sure if you're allowed to use the mean value theorem, but anyway, I'll write it down and wish it to be useful.

Let's define function $f$ over $(0,\infty)$ as $$f(x)=\mathrm{e}^{\frac1x}$$ Since $f$ is differentiable over its domain, according to the mean value theorem, for every $a, b \in (0,\infty)$ and $a\le b$, there exists $c\in [a,b]$ where $$f'(c)=\frac{f(b)-f(a)}{b-a}$$

Now, by substituting $a$ and $b$ with $x$ and $x+1$, respectively, and inserting the derivative of $f$, we'll get $$-\frac{1}{c^2}\mathrm{e}^{\frac1c}=\mathrm{e}^{\frac1{x+1}}-\mathrm{e}^{\frac1{x}},\quad c\in[x,x+1]$$ and note that since $c$ is dependent to $x$, one can rewrite the above equation as $$-\frac{1}{c^2(x)}\mathrm{e}^{\frac1{c(x)}}=\mathrm{e}^{\frac1{x+1}}-\mathrm{e}^{\frac1x},\quad c(x)\in[x,x+1]$$

Multiplying both sides by $-x^2$ results in $$\frac{x^2}{c^2(x)}\mathrm{e}^{\frac1{c(x)}}=x^2\left(\mathrm{e}^{\frac1x}-\mathrm{e}^{\frac1{x+1}}\right),\quad c(x)\in[x,x+1]$$

Now, the original limit in question, $L$, would be

\begin{aligned}L&=\lim_{x\to\infty}{x^2\left(\mathrm{e}^{\frac1x}-\mathrm{e}^{\frac1{x+1}}\right)}\\ &=\lim_{x\to\infty}{\frac{x^2}{c^2(x)}\mathrm{e}^{\frac1{c(x)}}},\quad c(x)\in[x,x+1] \end{aligned}

The above limit can be splitted into two limits, i.e., $$L=\lim_{x\to\infty}{\frac{x^2}{c^2(x)}}\lim_{x\to\infty}{\mathrm{e}^{\frac1{c(x)}}},\quad c(x)\in[x,x+1]$$

Both of the above limits are equal to unity. For the first limit, since $x\le c(x) \le x+1$, then $$\frac{x^2}{(x+1)^2} \le \frac{x^2}{c^2(x)}\le \frac{x^2}{x^2}=1$$ which by using the squeeze theorem results in $$L_1=\lim_{x\to\infty}{\frac{x^2}{c^2(x)}}=1$$ You can use a similar argument for the second limit and thus $$L_2=\lim_{x\to\infty}{\mathrm{e}^{\frac1{c(x)}}}=1$$ Finally $L$ will be found $$L=L_1L_2=1$$