Find the limit $L=\lim_{x\to0} \left(\frac{e^{-x}+x}{e^x-x}\right)^\frac{1}{x^2}$

Question

Find the limit $$L=\lim_{x\to0} \left(\dfrac{e^{-x}+x}{e^x-x}\right)^\frac{1}{x^2}$$

My try: We can write the function of which we are trying to find the limit as $x\to0$ as $$e^{\frac{1}{x^2}\ln\left(\dfrac{e^{-x}+x}{e^x-x}\right)},$$ so our answer would be $e^{L_1},$ where $$L_1=\lim_{x\to0}\dfrac{1}{x^2}\ln\left(\dfrac{e^{-x}+x}{e^x-x}\right)=\lim_{x\to0}\dfrac{\ln\left(\dfrac{e^{-x}+x}{e^x-x}\right)}{\frac{e^{-x}+x}{e^x-x}-1}\cdot\dfrac{\frac{e^{-x}+x}{e^x-x}-1}{x^2}=\lim_{x\to0}\dfrac{e^{-x}+2x-e^x}{(e^x-x)x^2}=\lim_{x\to0}\dfrac{-e^{-x}-e^x+2}{(e^x-1)x^2+2x(e^x-x)},$$ which is still $\frac{0}{0}$. Any ideas? It'll get messy if I apply L'Hopital one more time.

Answer 1

$\text{Altenatively, the Taylor series expansion can be used to solve $L_1$ very efficiently,}$ $$L_1{=\lim_{x\to0}\dfrac{1}{x^2}\ln\left(\dfrac{e^{-x}+x}{e^x-x}\right)=\lim_{x\to0}\dfrac{\ln\left(\dfrac{e^{-x}+x}{e^x-x}\right)}{\frac{e^{-x}+x}{e^x-x}-1}\cdot\dfrac{\frac{e^{-x}+x}{e^x-x}-1}{x^2}\\ =\lim_{x\to0}\dfrac{e^{-x}+2x-e^x}{(e^x-x)x^2} \\=\lim_{x\to0}\dfrac{e^{-x}+2x-e^x}{x^2}\cdot\lim_{x\to0}\dfrac{1}{(e^x-x)}=\lim_{x\to0}\dfrac{e^{-x}+2x-e^x}{x^2}\\=\lim_{x\to0}\dfrac{\left[1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} +O(x^4)\right]+2x-\left[1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +O(x^4)\right]}{x^2}\\ =\lim_{x\to0}\dfrac{-2\left[\frac{x^3}{3!}+\frac{x^5}{5!}+O(x^7)\right]}{x^2}\\=0}$$ $\text{Therefore,}$

$$L=e^{L_1}=e^0=1$$

Answer 2

In order to use L’Hospital Rule, we first take the logarithm of the limit as: $$ \begin{aligned} \lim _{x \rightarrow 0} \frac{\ln \left(\frac{e^{-x}+x}{e^x-x}\right)}{x^2} = & \lim _{x \rightarrow 0} \frac{\ln \left(e^{-x}+x\right)-\ln \left(e^x-x\right)}{x^2}\\ = & \frac{1}{2}\left[\lim _{x \rightarrow 0} \frac{e^x-1}{x\left(x e^x+1\right)}-\lim _{x \rightarrow 0} \frac{e^x-1}{x(e^x-x)}\right] \\ = & \frac{1}{2}\left[\lim _{x \rightarrow 0} \frac{e^x-1}{x} \cdot\left(\lim _{x \rightarrow 0} \frac{1}{x e^x+1}-\lim _{x \rightarrow 0} \frac{1}{e^x-x}\right)\right] \\ = & \frac{1}{2}[1 \cdot(1-1)] \cdots (*)\\ = & 0 \end{aligned} $$ where $(*)$ uses the result $\lim _{x \rightarrow 0} \frac{e^x-1}{x}=1$.

Hence $$ \boxed{L=e^0=1} $$

Answer 3

To avoid l'Hospital and Taylor expansion, we have

$$ \left(\dfrac{e^{-x}+x}{e^x-x}\right)^\frac{1}{x^2}= \left(1+\dfrac{2x+e^{-x}-e^x}{e^{x}-x}\right)^\frac{1}{x^2}=$$

$$=\left[\left(1+\dfrac{2x+e^{-x}-e^x}{e^{x}-x}\right)^{\dfrac{e^{x}-x}{2x+e^{-x}-e^x}}\right]^{\dfrac{2x+e^{-x}-e^x}{(e^{x}-x)x^2}}\to e^0=1$$

indeed by standard limits

$$\left(1+\dfrac{2x+e^{-x}-e^x}{e^{x}-x}\right)^{\dfrac{e^{x}-x}{2x+e^{-x}-e^x}}\to e$$

and using that $\frac{e^x-x-1}{x^2}\to L = \frac12$ (reference)

$$\dfrac{2x+e^{-x}-e^x}{(e^{x}-x)x^2}=\dfrac{(e^{-x}+x-1)-(e^x-x-1)}{(e^{x}-x)x^2}=$$

$$=\frac1{e^{x}-x}\left(\dfrac{e^{-x}+x-1}{x^2}-\dfrac{e^x-x-1}{x^2}\right)\to 1\cdot\left(\frac12-\frac12\right)=0$$

Answer 4

Actually, you don't need to do a lot of concrete calculations. Just realize that the expression you are interested in has the structure $$\frac 1{x^2}\left(g(x) - g(-x)\right) \text{ with } g(x) = \ln(e^{-x} +x)$$ Now, via Taylor you get $$g(\pm x) = g(0) \pm g'(0)x + \frac 12 g''(0) x^2 \pm \frac 16 g'''(0) x^3+ o(x^3)$$ You can quickly verify that $$g'(0) = 0$$ Hence, $$\frac 1{x^2}\left(g(x) - g(-x)\right) = \frac 1{x^2}\left(\frac 13 g'''(0)x^3+ o(x^3)\right)=\frac 13 g'''(0)x+ o(x)\stackrel{x\to 0}{\longrightarrow}0$$ So, the limit searched for ist $e^0 = 1$.