Find the limit of $\lim\limits_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}$ as $x$

Question

Find the limit of $\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}$ as $x$ approaches right of zero.

The answer is $\frac{1}{2}$ but I keep getting 1. Here's what I have:

Since $\lim_{x\to0^+} \frac{\ln (1+x)}{x}$ is of indeterminate form (0/0), we can apply L Hôpitals.

So now I have $\lim_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{1}{1+x}}{x}$, which is also of indeterminate form (0/0).

So by L Hôpitals, $\lim_{x\to0^+}\frac{1}{e} \frac{{- e^ \frac{1}{1+x} \frac{-1}{(1+x)^2}}}{x}$ which is equal to 1.

But as I said, the answer is $\frac12$. Can you tell me where did I go wrong? Thanks!

Answer 1

You can't apply l'Hopital rule in this way (note also that the derivative of $\frac{\ln (1+x)}{x}$ seems to be wrong)

$$\color{red}{\lim_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}= \lim_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{1}{1+x}}{x}}$$

you should derive all the terms as follow

$$\color{green}{\lim_{x\to0^+}\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}= \lim_{x\to0^+} \frac{- e^ \frac{\ln (1+x)}{x} (\frac{\ln (1+x)}{x})' }{e}}$$

but it seems that we don't get a better expression.

As an alternative note that

$$\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x}= \frac{1- e^{\frac{\ln (1+x)}{x}-1}}{\frac{\ln (1+x)}{x}-1}\frac{\frac{\ln (1+x)}{x}-1}{x}= \frac{e^{\frac{\ln (1+x)}{x}-1}-1}{\frac{\ln (1+x)}{x}-1}\frac{x- \ln (1+x) }{x^2}\to1\cdot\frac12=\frac12$$

indeed

$$\ln (1+x)=x-\frac{x^2}{2}+o(x^2)\implies \frac{x-\ln (1+x) }{x^2}=\frac{x-x+\frac{x^2}{2}+o(x^2) }{x^2}=\frac12+o(1)\to\frac12$$

Answer 2

$$\lim_{x \to 0}\frac{1}{e} \frac{e- e^{\dfrac{\ln (1+x)}{x}}}{x} = \lim_{x \to 0} \frac{1- e^{\dfrac{\ln (1+x)}{x} - 1} }{x} \\= \lim_{x \to 0} \frac{1- e^{\dfrac{\ln (1+x)}{x} - 1} }{\dfrac{\ln (1+x)}{x} - 1} \lim_{x \to 0} \dfrac{\dfrac{\ln (1+x)}{x} - 1}{x} =-\lim_{x\to 0} \dfrac{\ln(1 + x) - x}{x^2} = \dfrac12$$

Proof of the last limit.

Answer 3

Given that $$\ln (1+x)=x-\frac{x^2}{2} +o(x^2)$$ we have $$\frac{1}{e} \frac{e- e^ \frac{\ln (1+x)}{x}}{x} = \frac{1- e^ \frac{\ln (1+x)-x}{x}}{x} = \frac{1- e^{ \frac{-x}{2}+o(x)} }{x} =\frac{1- 1+\frac{x}{2}+o(x) }{x} = \frac{1}{2}+o(1)\to\frac{1}{2}$$

Answer 4

$$\lim_{x\rightarrow0}\frac{e- e^ \frac{\ln (1+x)}{x}}{ex}=-\lim_{x\rightarrow0}\frac{e^ \frac{\ln (1+x)}{x}\left(\frac{x}{1+x}-\ln(1+x)\right)}{ex^2}=-\lim_{x\rightarrow0}\frac{x-(1+x)\ln(1+x)}{x^2}=$$ $$=-\lim_{x\rightarrow0}\frac{1-\ln(1+x)-1}{2x}=\frac{1}{2}.$$