Solve limit with Lagrange theorem

Question

I tried to solve this limit: $$ \lim_{x \to +\infty} x^2\left(e^{\frac{1}{x+1}}-e^{\frac{1}{x}}\right) $$

Instead of solving it with Taylor series (using $u = 1/x$), I noticed that the difference within the parenthesis is the $\Delta f$ of the function $e^{\frac{1}{x}}$.

$\lim\limits_{x \to +\infty} x^2 * \Delta\left(e^{1/x}\right) = $

The Lagrange theorem says that $\Delta f = D\left[f\right] * \Delta x$, so

$=\lim\limits_{x \to +\infty} x^2 * \left( D\left[e^{1/x}\right]_{x_0} * \Delta x\right), x_0 \in \left(x, x+1\right)$

$$ x_0=x+r(x)$$ $$ 0<r(x)<1 $$

$=\lim\limits_{x \to +\infty} x^2 * \left( e^{\frac{1}{x+r(x)}} * \left(-\frac{1}{\left(x+r(x)\right)^2}\right)\right) $

$=\lim\limits_{x \to +\infty} -\frac{1}{\left(x+r(x)\right)^2} * x^2 * e^{\frac{1}{x+r(x)}}$

$\approx \lim\limits_{x \to +\infty} -\frac{1}{x^2} * x^2 * e^{\frac{1}{x}} $

$=\lim\limits_{x \to +\infty} -1 * e^{\frac{1}{x}} = -1$

Is it correct to proceed in this way?

EDIT: Using approximations is not a very elegant solution. I might be more precise using the Squeeze theorem (I'll think the solution, then I'll post it)

Instead of using the approximation, we consider these three functions:

• $f(x) = x^2 * e^{\frac{1}{x+1}} * \frac{1}{(x+1)^2}$
• $h(x) = x^2 * e^{\frac{1}{x+r(x)}} * \frac{1}{(x+r(x))^2}$
• $g(x) = x^2 * e^{\frac{1}{x}} * \frac{1}{x^2}$

Obviously, $f(x) \le h(x) \le g(x)$ for $x > 0$ and $r(x) \in (0, 1)$.

$\lim\limits_{x \to +\infty} f(x) = \lim\limits_{x \to +\infty} x^2 * e^{\frac{1}{x+1}} * \frac{1}{(x+1)^2} = 1$

(It is 1 and not -1 just because I left the minus sign out of the limit)

$\lim\limits_{x \to +\infty} g(x) = \lim\limits_{x \to +\infty} x^2 * e^{\frac{1}{x}} * \frac{1}{x^2} = 1$

So $\lim\limits_{x \to +\infty} h(x) = 1 \rightarrow \lim\limits_{x \to +\infty} -h(x) = \lim\limits_{x \to +\infty} x^2\left(e^{\frac{1}{x+1}}-e^{\frac{1}{x}}\right) = -1$

Here's how the f (the red function) and g (the blue function) look like:

Answer 1

We can also proceed in the following manner \begin{align} L &= \lim_{x \to \infty}x^{2}\{e^{1/(x + 1)} - e^{1/x}\}\notag\\ &= \lim_{x \to \infty}x^{2}e^{1/x}\{e^{1/(x + 1) - 1/x} - 1\}\notag\\ &= \lim_{x \to \infty}x^{2}e^{1/x}\cdot\frac{e^{1/(x + 1) - 1/x} - 1}{1/(x + 1) - 1/x}\cdot\left(\frac{1}{x + 1} - \frac{1}{x}\right)\notag\\ &= -\lim_{x \to \infty}x^{2}e^{1/x}\cdot 1\cdot\frac{1}{x(x + 1)}\notag\\ &= -\lim_{x \to \infty}1\cdot\frac{x}{x + 1}\notag\\ &= -\lim_{x \to \infty}\dfrac{1}{1 + \dfrac{1}{x}}\notag\\ &= -1\notag \end{align} Your approach starts off correctly, but there is a problem in the middle when you use $\approx$ symbol. A limit evaluation is an exact operation and not an approximate one. But this is a minor error and you can just say that $e^{1/(x + r(x))} \to 1$ as $x \to \infty$ and the ratio $x^{2}/(x + r(x))^{2} \to 1$ as $x \to \infty$ (I hope you can justify easily why both these limits should be $1$).

Answer 2

\begin{align} & \lim_{x \rightarrow \infty} x^2 \left( e^{\frac{1}{x+1}} -e^{\frac{1}{x}} \right)\\ = & \lim_{x \rightarrow \infty} \left( 1+ \frac{1}{x+1} + \frac{1}{2 (x+1)^2} + O(x^{-3}) - 1 -\frac{1}{x} - \frac{1}{2 x^2} + O(x^{-3})\right) \\ = & \lim_{x \rightarrow \infty} \frac{-x^2}{(x+1)x} + \frac{-2x-1}{2(x+1)^2 x^2} +O(x^{-1}) = -1. \end{align}

Answer 3

I'll assume throughout that $x>0$.

By Lagrange’s theorem, you can write $$ \frac{f(x+1)-f(x)}{(x+1)-x}=f'(c_x) $$ for some $c_x\in (x,x+1)$. Thus $$ x^2(f(x+1)-f(x))=x^2f'(c_x)=-\frac{x^2}{c_x^2}f(c_x) $$ Now $$ f(x+1)\le f(c_x)\le f(x),\qquad x^2\le c_x^2\le (x+1)^2 $$ and putting the inequalities together $$ \frac{x^2}{(x+1)^2}f(x+1)\le \frac{x^2}{c_x^2}f(c_x)\le f(x) $$ that can be rewritten as $$ -f(x)\le x^2(f(x+1)-f(x))\le -\frac{x^2}{(x+1)^2}f(x+1) $$ Since $\lim_{x\to\infty}f(x)=1$ we conclude by the squeeze theorem that your limit is $-1$.