How to evaluate the limit $\lim\limits_{x\to\infty}\Bigl(1+2^x\Bigr)^{\frac{1}{x}}$? I stumbled across this limit question and i'm not sure of how to handle this.

$\lim\limits_{x\to\infty}\Bigl(1+2^x\Bigr)^{\frac{1}{x}}$

I think this limit goes to 1 because $\lim\limits_{x\to\infty}\frac{1}{x}=0$. Is this correct way to solve this limit?
 A: The function $\ln$ is continuous in the interval $(0,\infty)$, so
$$\ln\lim_{x\to\infty}(1+2^x)^{1/x}=\lim_{x\to\infty}\frac{\ln(1+2^x)}x=\lim_{x\to\infty}\frac{2^x\ln 2}{1+2^x}=\ln 2$$
Then
$$\lim_{x\to\infty}(1+2^x)^{1/x}=2$$
A: Let $k>2$ be an arbitrary real number. Then for $x$ large enough*, we have
$$
2^x\leq 1+2^x\leq k^x
$$
Now take the $x$th root of all of them, giving
$$
2\leq \left(1 + 2^x\right)^{1/x}\leq k
$$
So for any $k>2$, eventually $\left(1 + 2^x\right)^{1/x}$ lies between $2$ and $k$. In other words, we have that for any $k>2$, there is a real number $X$ such that
$$
\left|2- \lim_{x\to\infty}\left(1 + 2^x\right)^{1/x}\right|\leq k-2
$$
for any $x>X$, showing that the limit exists and is equal to $2$.

*We have
$$
1+2^x\leq k^x\\
2^{-x} + 1\leq \left(\frac k2\right)^x
$$
Any positive $x$ that makes $2<\left(\frac k2\right)^x$ must fulfill the above inequality. We get
$$
\ln 2<x(\ln k-\ln 2)\\
\frac{\ln 2}{\ln k - \ln 2}< x
$$
so with $X = \frac{\ln 2}{\ln k - \ln 2}$ any $x>\max(X, 0)$ will definitely work.
A: No, it certainly is not correct, as it lacks of rigour. $\infty^0$ is an indeterminate form.
You can proceed like this: determine the limit of the log first and use equivalence of functions:
$$\ln(1+2x)^{\tfrac 1x}=\frac{\ln(1+2^x)}x\sim_\infty\frac{\ln(2^x)}x=\frac{\not x\ln2}{\not x}=\ln 2.$$
Indeed, $\;1+2^x\sim_\infty 2^x$, hence $\ln(1+2^x)\sim_\infty \ln(2^x)$.
A: Your explanation is wrong. $\dfrac1x$ does tend to $0$, but you ignore the fact that $x$ goes to infinity at the same time. (You might as well have said the limit is infinity because $x$ goes to infinity.)
A correct way is to decompose 
$$(1+2^x)^{1/x}=2\left(\frac1{2^x}+1\right)^{1/x}.$$
Now the quantity between parenthesis tends to $1$, and so does its $x^{th}$ root.
