Calculating the limit of fractions for uniform convergence I'm trying to calculate the limit sup of a term and see if it equals zero in order to prove uniform convergence. I arrive at this:
$\lim_{n\to \infty} \sup_{x \in [0,1]} |f_{n}-f(x)| = $
$\lim_{n\to \infty}(1-\frac{n}{n+2})(\frac{n}{n+2})^\frac{n}{2}$
now how I solved the limit is such:
$\lim_{n\to \infty}(1-\frac{n}{n+2})(\frac{n}{n+2})^\frac{n}{2} $
$=\lim_{n\to \infty}(1-\frac{n}{n(1+2/n)})(\frac{n}{n(1+2/n)})^\frac{n}{2}$
the first term equals 1-1 since 2/n converges zero. hence the limit is qual zero, which is also what is said in the solution. However they calculated the limit in a completely other way, one where you get terms such as e^-1, but the limit still equals zero. My qustion is: is my method correct?
here how it's solved in the textbook:
$\lim_{n\to \infty}(1-\frac{n}{n+2})(\frac{n}{n+2})^\frac{n}{2} $
$=\lim_{n\to \infty}(1-\frac{n}{n+2})(1-\frac{2}{n+2})^\frac{n}{2} $
$=\lim_{n\to \infty}(1-\frac{n}{n+2})\frac{1}{e} = 0 $
I do not understand the last line of this solution at all. Any help would be greatly appreciated.
Thank you for your help

Hint: We have $(1 + \frac{x}{n})^n \rightarrow \exp(x)$
