Find the limit $\lim_{x\to\infty} \left(e^{\frac{1}{x}}+\frac{1}{x}\right)^{x}$

Question

$\lim_{x\to\infty} \left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)^{x}$

Can you help me to find this limit. I am unable to find a way to finish this beast. Even though I have done some similar limits involving $e$ number.

Should I take the logarithm and use L'Hopital rule?

What's stopping you from doing what you suggest and use l'Hopital? — Arthur, Oct 23 '19 at 23:31
At the very least I think you should try. You have an idea for an approach that might work or it might not work. You shouldn't need us to tell you to just go ahead. — Arthur, Oct 23 '19 at 23:53
The approach $t=\frac1x\leadsto \log\leadsto$ L'Hospital works certainly. — A.Γ., Oct 24 '19 at 00:14
Does this answer your question? Limit of a Function: $ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}$ — Anne Bauval, Jun 01 '23 at 14:16

Nanayajitzuki · Answer 1 · 2019-10-24T01:49:53.960

simply thinking $$e^x = 1 + x + O(x^2) \quad (x\to0)$$ so $$\left(e^{\frac1{x}}+\frac1{x}\right)^x =\left(1+\frac2{x}+O(x^{-2})\right)^x\to e^2 \quad (x\to\infty)$$

edit for some supplement, in general if you meet $(1+x+o(x))^{\frac1{x}}$ type limitation for $x\to0$ you always have $$\frac1{x}\ln(1+x+o(x))=\frac1{x}(x+o(x)+o(x+o(x)))\to1$$ hence you will produce $e$ in result. in your case, you write $$\left(1+\frac2{x}+O(x^{-2})\right)^x=\left(1+\frac2{x}+O(x^{-2})\right)^{\frac{x}{2}\cdot2}$$ leads to $e^2$

moreover $(1+f(x)+o(f(x)))^{\frac1{g(x)}}$ always leads to $e^z$ and $z$ is almostly decided by the limitation of $\frac{f(x)}{g(x)}$ (not that strict, but you always have $\frac1{g(x)}=\frac1{f(x)}\frac{f(x)}{g(x)}$ on shoulder), so you may deal with it by this fashion, what i write before is a rough thinking to get the result.

@James Warthington if you have no problem with the negligible of item $O(x^2)$ in $(1+x)^\frac1{x}$ type limitation, i think the result is obvious. if not you may need take $\ln$ to prove it strictly — Nanayajitzuki, Oct 23 '19 at 23:50

score 3 · Answer 2 · answered Oct 23 '19 at 23:45

3

Considering $$f(x)=\left(e^{1/x}+\frac1x\right)^x$$ we have that $$\begin{align} \lim_{x\to\infty}\ln{(f(x))} &=\lim_{x\to\infty}x\ln{\left(e^{1/x}+\frac1x\right)}\\ &=\lim_{x\to\infty}x\left(\ln{\left(e^{1/x}\right)}+\ln{\left(1+\frac1{xe^{1/x}}\right)}\right)\\ &=\lim_{x\to\infty}\left(1+x\ln{\left(1+\frac1{xe^{1/x}}\right)}\right)\\ &=\lim_{x\to\infty}\left(1+x\left(\frac1{xe^{1/x}}+o\left(\frac1{xe^{1/x}}\right)\right)\right)\\ &=\lim_{x\to\infty}\left(1+\frac1{e^{1/x}}+o\left(\frac1{e^{1/x}}\right)\right)\\ &=2\\ \end{align}$$ Hence the limit is given by $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}e^{\ln{(f(x))}}=e^{\lim_{x\to\infty}\ln{(f(x))}}=e^2$$

answered Oct 23 '19 at 23:45

Peter Foreman

19,947

@SimonFraser What part exactly? Can I make it clearer? – Peter Foreman Oct 23 '19 at 23:49
I was just kidding. I didn't downvote it. Anyways, does $o(\frac{1}{xe^{1/x}})$ refer to an error function? – Simon Fraser Oct 23 '19 at 23:57
Yes it is Little-o notation. I used that $\ln{(1+x)}=x+o(x)$ as $x\to0$. – Peter Foreman Oct 24 '19 at 00:01

score 2 · Answer 3 · answered Oct 24 '19 at 02:00

2

Replace $x$ by ${1 \over x}$. Your limit becomes $$\lim_{x \rightarrow 0^+} (e^x + x)^{1 \over x}$$ Taking logarithms, the natural log of your limit is $$\lim_{x \rightarrow 0^+} {\ln(e^x + x) \over x}$$ Now you can use l'hopital's rule to get that this is the same as $$\lim_{x \rightarrow 0^+} {e^x + 1 \over e^x + x}$$ Plug in $0$ to get the limit being $2$. This is the natural log of the original limit, so your original limit is $e^2$.

answered Oct 24 '19 at 02:00

Zarrax

44,950

You mean $t$ instead of $x$? – James Warthington Oct 24 '19 at 02:12
Very frequently, L'Hospital's rule is overkill and may be replaced by the more elementary recognition and calculation of some derivative. This is the case here: setting $$f(x)=\ln(e^x+x),$$ we find immediately $$\lim_{x\rightarrow 0} \frac{\ln(e^x+x)}x=f'(0)=2.$$ – Anne Bauval Jun 01 '23 at 14:12

score 1 · Answer 4 · answered Oct 24 '19 at 01:41

Letting $u=1/x\to0^+$ and using the limit theorem $\lim(f^g)=(\lim f)^{\lim g}$ if $\lim f$ and $\lim g$ both exist and aren't both equal to $0$ (and $f$ avoids negative values, so that $f^g$ is well defined), we have

$$\left(e^{1/x}+{1\over x} \right)^x=(e^u+u)^{1/u}=e\left(\left(1+{u\over e^u}\right)^{e^u/u}\right)^{1/e^u}\to e(e)^1=e^2$$

since $(1+1/w)^w\to e$ as $w=e^u/u\to\infty$.

James Warthington · Answer 5 · 2019-10-24T02:07:17.283

0

I wish to try this:

$\lim_{x\to\infty} \left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)^{x}= \lim_{x\to\infty} e^{\ln\left(e{\frac{1}{x}}+\dfrac{1}{x}\right)^{x}}= \lim_{x\to\infty}e^{x\ln\left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)}=\lim_{x\to\infty}e^\dfrac{{\ln\left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)}}{\dfrac{1}{x}}=\lim_{x\to\infty}e{\dfrac{\dfrac{d}{dx}\ln\left(e^{\frac{1}{x}}+\dfrac{1}{x}\right)}{\dfrac{d}{dx}(\dfrac{1}{x})}}=\lim_{x\to\infty}e^\left(\dfrac{\dfrac{-e^{\frac{1}{x}}-1}{x(xe^{\frac{1}{x}}+1)}}{-\dfrac{1}{x^2}}\right)=\lim_{x\to\infty}e^\left(\dfrac{-x^{2}(-e^{\frac{1}{x}}-1)}{x(xe^{\frac{1}{x}}+1)}\right)=\lim_{x\to\infty}e^\left(\dfrac{-x(-e^{\frac{1}{x}}-1)}{xe^{\frac{1}{x}}+1}\right)=\lim_{x\to\infty}e^\left(\dfrac{\dfrac{-x(-e^{\frac{1}{x}}-1)}{x}}{\dfrac{xe^{\frac{1}{x}}+1}{x}}\right)=\lim_{x\to\infty}e^\left(\dfrac{e^{\frac{1}{x}}+1)}{{e^{\frac{1}{x}}+\dfrac{1}{x}}}\right)=e^{\dfrac{1+1}{1+0}}=e^2$

The answer seems to be $e^{2}$, so I must make some mistakes along the way. This is the most challenging post I have ever posted using latex.

Edit: THanks to A.Γ., I have been able to correct all steps.

edited Oct 24 '19 at 02:07

answered Oct 24 '19 at 00:28

James Warthington

3,187

Did you see my comment above to consider the variable change $t=\frac1x$? – A.Γ. Oct 24 '19 at 00:34
@A.Γ. Can you do this in a separate answer post? Your comment is too brief so I don't grasp the whole picture. – James Warthington Oct 24 '19 at 00:35
@A.Γ. The answer is supposed to be $e^2$, mine is 1. Where did I commit a mistake, can you have a look? – James Warthington Oct 24 '19 at 00:38
After the variable change you have to calculate $$\lim_{t\to 0}\exp{\frac{\ln(e^t+t)}{t}}=\exp\left({\lim_{t\to 0}\frac{\ln(e^t+t)}{t}}\right).$$ Apply L'Hospital. – A.Γ. Oct 24 '19 at 00:50
This limit diverges right? – James Warthington Oct 24 '19 at 00:56
Do you mean $$\exp(\lim_{t\to\infty} \dfrac{\ln(e^{t}+1)}{t})$$? This approaches $e$ – James Warthington Oct 24 '19 at 01:05
The limit converges to $2$. I mean exactly the limit in my comment. Your calculations in the post are still erroneous. Check out the numerator signs after you multiply it be the negative. – A.Γ. Oct 24 '19 at 01:37
@A.Γ. Why approaches 0, I don't understand this point – James Warthington Oct 24 '19 at 01:42
Well, if $x\to+\infty$ what can you say about $t=\frac{1}{x}$? – A.Γ. Oct 24 '19 at 01:51
@A.Γ. I see now, thank you for your patient guidance. I have corrected my answer and all seem good. Thank you again! – James Warthington Oct 24 '19 at 02:08

score 0 · Answer 6 · answered Oct 24 '19 at 21:21

0

Let $u=\frac1x$ and then \begin{eqnarray} &&\lim_{x\to\infty}x\ln(e^{\frac1x}+\frac1x)\\ &=&\lim_{u\to0}\frac{\ln(e^u+u)}{u}=\lim_{u\to0}\frac{\ln[1-(1-e^u-u)]}{u}\\ &=&\lim_{u\to0}\frac{\ln[1-(1-e^u-u)]}{1-e^u-u}\cdot\frac{1-e^u-u}{u}\\ &=&(-1)(-2)=2 \end{eqnarray} and hence $$ \lim_{x\to\infty}(e^{\frac1x}+\frac1x)^x=e^2.$$ Here $$ \lim_{x\to0}\frac{\ln(1-x)}{x}=\lim_{x\to0}\frac{1-e^x}{x}=-1$$ are used.

answered Oct 24 '19 at 21:21

xpaul

44,000

1

A slightly simpler construction: $$\frac{\ln(e^u+u)}{u}=\frac{\ln(e^u(1+ue^{-u})}{u}=\frac{u+\ln(1+ue^{-u})}{u}=1+\frac{\ln(1+ue^{-u})}{ue^{-u}}e^{-u}\to 1+1\cdot 1=2.$$ – A.Γ. Oct 26 '19 at 07:43
@A.Γ.，you are right. Thanks. – xpaul Oct 27 '19 at 00:28

Find the limit $\lim_{x\to\infty} \left(e^{\frac{1}{x}}+\frac{1}{x}\right)^{x}$

6 Answers6