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!

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
$$\bigg({4 + \frac{4}{n} \bigg)^{\frac{n}{2}}}\ge 4^{\frac n2} \to \infty$$
