Issue with the following limit $\bigg(2 \sqrt{1 + \frac{1}{n}}\bigg)^n$ Calculate the following limit:
$\bigg(2  \sqrt{1 + \frac{1}{n}}\bigg)^n$
When I calculate it I get to different answers.
First way (Edit: this is where I did the mistake): $$\bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n = \bigg({4 + \frac{4}{n} \bigg)^\frac{n}{2}} = \bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4} \cdot \frac{4}{n}\cdot\frac{n}{2}}}$$
When we do $\lim_{n \to \infty}\bigg(\bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{4}\cdot \frac{4}{n}\cdot\frac{n}{2}}}\bigg)$  we get $e^2$
Now the second way:
$$\bigg(2 \cdot \sqrt{1 + \frac{1}{n}}\bigg)^n = 2^n\cdot (1 + \frac{1}{n})^{n \cdot \frac{1}{2}}$$
When we do limit out of this we get $2^\infty \cdot \sqrt{e}$ which is of course $\infty$.
Could someone point out the mistake I made?
Edit:
I just realised where my mistake lies! I mistakenly thought that $(4 + \frac{4}{n})^\frac{n}{4} = e$ which is false, actually $(1 + \frac{4}{n})^\frac{n}{4} = e$. The second way of calculating this limit is the correct one!
 A: Your use of limit for $e^x$ is incorrect.  
$$\lim\limits_{n\to\infty}\left(1+\dfrac{4}{n}\right)^n = e^4$$ 
$$\lim\limits_{n\to\infty}\left(4+\dfrac{4}{n}\right)^n = \infty$$
A: What about using the limit chain rule.
$$\lim_{n\to \infty} \left(4+\frac{4}{n}\right)^{\frac{n}{2}} =\lim_{n\to \infty} e^{\frac{n}{2}\ln\left(4+\frac{4}{n}\right)}$$
Here you can see why the limit diverges (how does $\frac{n}{2}\ln\left( 4+\frac{4}{n}\right)$ behave as $n$ tends to infinity), as opposed to something like $\left(1+\frac{a}{n}\right)^{\frac{n}{2}}$.
A: $$
\lim\limits_{x\to a} f(x)g(x) \ne \lim\limits_{x\to a} f(x).\lim\limits_{x\to a} g(x)
$$
if the individual limits don't exist
Therefore, to say
$$
\lim\limits_{n\to \infty} \bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n = \lim\limits_{n\to \infty} 2^n * \lim\limits_{n\to \infty}(1 + \frac{1}{n})^{\frac{n}{2}}
$$
is incorrect since the first limit on the RHS doesn't exist.
The second way is therefore inapplicable. @user has given the excellent observation which gives the answer
$$
\bigg(2 * \sqrt{1 + \frac{1}{n}}\bigg)^n\ge 2^n \to \infty
$$
A: Since$$\lim_{n\to\infty}4+\frac4n=4\text{ and }\lim_{n\to\infty}\frac n4\times\frac4n\times\frac n2=\infty,$$we also have$$\lim_{n\to\infty}\left(4+\frac4n\right)^{\frac n4\times\frac4n\times\frac n2}=\infty,$$instead of $e^2$.
A: We have that
$$\bigg(2  \sqrt{1 + \frac{1}{n}}\bigg)^n=2^n\cdot \sqrt{\left(1+\frac1n\right)^n} \to \infty\cdot\sqrt e$$
and by your first method
$$\bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{2}}}\ge 4^{\frac n2} \to \infty$$
