Need help solving a limit (e^x + x)^(1/x)

[Whiplash]

The problem is lim as x -> infinity, (e^x + x)^(1/x)

Steps I did so far:
Asked the teacher and he said to use natural log
ln y = (1/x)ln(e^x + x)
ln y = ln(e^x + x)/x
ln y = (1/e^x + x) / 1 <- Took the derivative of step 2
ln y = 1/(e^x + x)
limit as x -> infinity, 1/(e^x + x)
I dont know where to go now. I used the limit on my calculator and it says the limit is e. So the above must evaluate to 1. Any help?

 

[galactus]

We have :

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

=\(\displaystyle \L\\\lim_{x\to\infty}e^{\frac{ln(e^{x}+x)}{x}}\)

=\(\displaystyle \L\\e^{\lim_{x\to\infty}\frac{ln(e^{x}+x)}{x}}\)

You already ascertained that the limit in the exponent is 1

Therefore, you have \(\displaystyle e^{1}\) as the limit.

 

[soroban]

Hello, Whiplash!

Your differentiation is off . . .

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

Let \(\displaystyle y \:=\:\left(e^x\,+\,x\right)^{\frac{1}{x}}\)

Then: \(\displaystyle \:\ln(y)\:=\:\ln\left[\left(e^x\,+\,x\right)^{\frac{1}{x}}\right] \:=\:\frac{1}{x}\ln\left(e^x\,+\,x\right)\)

We have: \(\displaystyle \:\lim_{x\to\infty}\,\frac{\ln\left(e^x\,+\,x\right)}{x}\) . . . which goes to \(\displaystyle \frac{\infty}{\infty}\)

Use L'Hopital: \(\displaystyle \L\:\lim_{x\to\infty}\,\frac{\frac{e^x\,+\,1}{e^x\,+\,x}}{1} \;=\;\lim_{x\to\infty}\,\frac{e^x\,+\,1}{e^x\,+\,x}\)

Divide top and bottom by \(\displaystyle e^x:\;\;\)\(\displaystyle \L\lim_{x\to\infty}\,\frac{1\,+\,\frac{1}{e^x}}{1\,+\,\frac{x}{e^x}}\;=\;\frac{1\,+\,0}{1\,+\,0} \;=\;1\)

We have:\(\displaystyle \L\:\lim_{x\to\infty}\left[\ln(y)\right]\:=\:1\;\;\Rightarrow\;\;\lim_{x\to\infty} y \:=\:e\)

Therefore: \(\displaystyle \L\:\lim_{x\to\infty}\left(e^x\,+\,x\right)^{\frac{1}{x}}\:=\:e\)