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!

The question as well as all the answers were downvoted. The hard work of the downvoter has been undone. — Paramanand Singh, Feb 01 '18 at 16:06
Please remember that you can choose an aswer among the given is the OP is solved, more details here https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work — user, Mar 09 '18 at 22:24

score 4 · Accepted Answer · edited Feb 01 '18 at 16:24

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

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

user8277998 · Answer 2 · 2018-02-01T16:13:17.697

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.

edited Feb 01 '18 at 16:13

answered Feb 01 '18 at 15:55

user8277998

2,666

Guy Fsone · Answer 3 · 2018-02-01T17:36:20.610

2

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

edited Feb 01 '18 at 17:36

answered Feb 01 '18 at 15:56

Guy Fsone

23,903

it is $\frac{1}{2}+o(1)$ and not $o(x)$ – stity Feb 01 '18 at 16:22
@stity yes sure thanks – Guy Fsone Feb 01 '18 at 17:38

score 1 · Answer 4 · answered Feb 01 '18 at 15:57

1

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

answered Feb 01 '18 at 15:57

Michael Rozenberg

194,933

Can you explain how you get from the second equality to the third? – Mashed Potato Feb 01 '18 at 16:09
It's L'Hospital and $(x-(x+1)\ln(1+x))'=1-1\cdot\ln(1+x)-(1+x)\cdot\frac{1}{1+x}.$ – Michael Rozenberg Feb 01 '18 at 16:11
How did $e$ get cancelled? – Mashed Potato Feb 01 '18 at 16:23
Because since $\frac{\ln(1+x)}{x}\rightarrow1$, we see that $e^{\frac{\ln(1+x)}{x}}\rightarrow e$. – Michael Rozenberg Feb 01 '18 at 16:25
How did you obtain $x- (1+x) ln (1+x)$? – Mashed Potato Feb 01 '18 at 16:32
Shouldn't there be a $(1+x)$ in the denominator? – Mashed Potato Feb 01 '18 at 16:37
$\frac{x}{1+x}-\ln(1+x)=\frac{x-(1+x)\ln(1+x)}{1+x}.$ Also, $1+x\rightarrow1$. – Michael Rozenberg Feb 01 '18 at 16:38
Did you already take the limit of the $(1+x)$ that's why it's missing in the expression? – Mashed Potato Feb 01 '18 at 16:40
The limit is equal to $1$ and I can miss it. It's just $\lim\limits_{x\rightarrow a}(f(x)g(x))=\lim\limits_{x\rightarrow a}f(x)\lim\limits_{x\rightarrow a}g(x)$, where two last limits they are exist. – Michael Rozenberg Feb 01 '18 at 16:42
Okay thank you so much!! – Mashed Potato Feb 01 '18 at 16:43
You are welcome! – Michael Rozenberg Feb 01 '18 at 16:44

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

4 Answers4