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

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!

• The intuitive answer is immediate: $x$ is quite negligible in front of $e^x$ and the expression tends to $e$.
– user65203
Apr 29, 2019 at 6:57

... just a silly mistake. If $$\ln L=1$$ , then what do you think $$L$$ should be equal to?

• So can I just do $e^1$, is this how l’hospitals rule normally works?
– user635953
Apr 25, 2019 at 15:27
• Oh is it because I took logs first then I have to reverse it?
– user635953
Apr 25, 2019 at 15:28
• Yes, first convert into $\frac 0 0$ or $\frac \infty \infty$ form, then differentiate until you get a determinate form. Apr 25, 2019 at 15:36
• Yes, you have to take log both sides, and then reverse it. What did you think initially!!!! Apr 25, 2019 at 15:37
• 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 $e^1=e$.
– user635953
Apr 25, 2019 at 15:52

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$$e^xand that therefore$$e<(e^x+x)^{\frac1x}<2^{\frac1x}e$$and so, by the squeeze theorem, your limit is $$e$$.

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

• You should justify this equivalence, shouldn't you ?
– user65203
Apr 29, 2019 at 6:59