Determine limit of indeterminate form.

Question

If the question is $$\lim_{x\to\infty}(e^x+1)^{\frac1x}$$ Do you just say that because $\lim_{x\to\infty}\frac1x$ is $0$, the original function has limit approaching 1, without caring the $e^x$?

Answer 1

Write: $$ (e^x+1)^{1/x}=(e^x)^{1/x} (1+e^{-x})^{1/x}=e\times (1+e^{-x})^{1/x} $$ which goes to $e$ as $x \to \infty$

Answer 2

Apply the squeeze theorem:

$$\lim_{x\to\infty}(e^x)^{\frac1x}\leq \lim_{x\to\infty}(e^x+1)^{\frac1x}\leq \lim_{x\to\infty}(e^x\cdot e)^{\frac1x}$$