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!