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I have this limit:

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

At first I was stumped but then decided to use L'hospitals rule and logs so it turns to:

$$\lim_{x\to \infty} \frac{\ln(e^x+x)}{x}$$

Then differentiating it twice turns to:

$$\lim_{x\to \infty} \frac{e^x}{e^x+1}$$

But then this means $\lim_{x\to \infty} \frac{e^x}{e^x+1}=1$, but I know from trying values on my calculator that it should be equal to $e$.

Am I wrong or am I getting mixed up with the L'Hospitals rule? Thank you!

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  • $\begingroup$ The intuitive answer is immediate: $x$ is quite negligible in front of $e^x$ and the expression tends to $e$. $\endgroup$ – Yves Daoust Apr 29 at 6:57
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... just a silly mistake. If $\ln L=1$ , then what do you think $L$ should be equal to?

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  • $\begingroup$ So can I just do $e^1$, is this how l’hospitals rule normally works? $\endgroup$ – Olly Reynolds Apr 25 at 15:27
  • $\begingroup$ Oh is it because I took logs first then I have to reverse it? $\endgroup$ – Olly Reynolds Apr 25 at 15:28
  • $\begingroup$ Yes, first convert into $\frac 0 0 $ or $\frac \infty \infty $ form, then differentiate until you get a determinate form. $\endgroup$ – Tojrah Apr 25 at 15:36
  • $\begingroup$ Yes, you have to take log both sides, and then reverse it. What did you think initially!!!! $\endgroup$ – Tojrah Apr 25 at 15:37
  • $\begingroup$ I was just confused why the limit was not $1$ but then I realised that I need to take the "exponential" of each side so remove the log so it leads to $e^1=e$. $\endgroup$ – Olly Reynolds Apr 25 at 15:52
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Why differentiate twice? After differentiating once you get$$\lim_{x\to\infty}\frac{e^x+1}{e^x+x}=\lim_{x\to\infty}\frac{1+e^{-x}}{1+\frac x{e^x}}=1$$and so, yes, your limit is equal to $e^1=e$.


Actually, you don't need L'Hopital's rule at all. Just note that$$e^x<e^x+x<2e^x$$and that therefore$$e<(e^x+x)^{\frac1x}<2^{\frac1x}e$$and so, by the squeeze theorem, your limit is $e$.

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You can also factor $e^x$ out like this: $$\lim_{x\to\infty} e \left(1+x e^{-x}\right)^{1/x}\sim \lim_{x\to\infty} e\left(1+e^{-x}\right)= e$$

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  • $\begingroup$ You should justify this equivalence, shouldn't you ? $\endgroup$ – Yves Daoust Apr 29 at 6:59

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