Solve $\lim_{x\to 0}\ (e^x+x)^{1/x}$ with L'Hopital's rule, not standard limit

Question

$$\lim_{x\to 0}\ (e^x+x)^{1/x}$$ I need to solve it using L'Hopital's rule.
I suppose, answer is $e^2$, but I'm not sure. Thank you.

It's ok just keep it the way it is... you got two answers one for $x\rightarrow 1$ and one for $x\rightarrow 0$. — Yanko, Dec 07 '19 at 16:27
Why would you impose the use of L'Hopital's rule, if some more elementary solution is available? + Quick beginner guide for asking a well-received question + please avoid "no clue" questions. — Anne Bauval, Jun 01 '23 at 13:48
Duplicate: Limit of a Function: $ \lim_{x \to 0}\ (e^x + x)^ {\large \frac {1} {x}}$. — Anne Bauval, Jun 01 '23 at 13:55

score 3 · Answer 1 · answered Dec 07 '19 at 16:25

3

You can't use L'Hospital here as this is not one of the forms that allow it:

$$(e^x+x)^{1/x}=e^{\frac1x\log(e^x+x)}\xrightarrow[x\to1]{}e^{\frac11\log(e+1)}=e+1$$

answered Dec 07 '19 at 16:25

DonAntonio

I think it should be $$e^{\frac1x\log(e^x+x)}\xrightarrow[x\to1]{}e^{\frac11\frac{1+1}{1}}=e^2$$ – user Dec 07 '19 at 16:45
@user I can't see why: $;e^x+x\to e+1;$ if $;x\to 1;$ ...You seem to have deleted the logarithm in the exponent... – DonAntonio Dec 07 '19 at 16:47
I was assuming you have applied l'Hospital. – user Dec 07 '19 at 16:52
1

@user Not at all. I even wrote that L'Hospital cannot be used here. – DonAntonio Dec 07 '19 at 17:19
But the limit is as $x\to 0$. – user Dec 07 '19 at 17:22
1

@user As you can read in the comments, it was not so at first. In fact, it was $;x\to 1;$ a good time after the question was posted and I answered. It changed afterwards. – DonAntonio Dec 07 '19 at 17:32
Ah ok sorry I didn’t noticed all that! Bye – user Dec 07 '19 at 17:33

score 1 · Accepted Answer · answered Dec 07 '19 at 16:26

1

If the limit is as $x\rightarrow 0$ (I see you edit) then you can use the fact that $\ln$ and $\exp$ are continuous to get that

$$\lim_{x\to 0}\ (e^x+x)^\dfrac{1}{x} = L$$

Iff

$$ \lim_{x\rightarrow 0} \frac{\ln(e^x+x)}{x} = \ln (L)$$

Now you can use the l'hopital rule

$$\lim_{x\rightarrow 0} \frac{\ln(e^x+x)}{x}=\lim_{x\rightarrow 0} \frac{e^x+1}{e^x+x}=\frac{2}{1}=2$$

Therefore $\ln(L)=2$ and so $L=e^2$.

answered Dec 07 '19 at 16:26

Yanko

Thank you. It's true – Александр Лисов Dec 07 '19 at 16:30
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 13:48

2 Answers2