Here is a useful equation:
$\displaystyle \displaystyle \lim_{x\to\infty} \left ( 1 + \dfrac{r}{x} \right ) ^{x} = e^r$.
For your problem, instead of evaluating the limit of the ratio, evaluate the ratio of the limits:
$\displaystyle \displaystyle \lim_{n\to\infty} \dfrac{ \left ( 1 + \dfrac{2}{n+1} \right ) ^{n+1} }{1 + \dfrac{2}{n+1}} = \dfrac{ \displaystyle \lim_{n\to\infty} \left ( 1 + \dfrac{2}{n+1} \right ) ^{n+1} }{ \displaystyle \lim_{n\to\infty} \left ( {1 + \dfrac{2}{n+1}} \right ) }$
The limit in the denominator is 1. Division by 1 won't change anything, so the answer to your original problem is just the limit in the numerator.
$\displaystyle \displaystyle \lim_{n\to\infty} \left ( 1 + \dfrac{2}{n+1} \right ) ^{n+1}$
If we let r = 2 and x = n + 1, then the limit in your problem becomes $\displaystyle \displaystyle \lim_{x\to\infty} \left ( 1 + \dfrac{r}{x} \right ) ^{x}$. Remember the useful equation at the top?
The limit is just e^r. Substituting r = 2 gives: e^2.
Thank you so much , it makes sense now.
My mistake was not evaluating the numerator correctly. I found the limit of the denominator correctly, but for the numerator I said that the limit when n approaches infinity was = (1)^ infinity which is the same as 1. What was wrong with this reasoning?
$\displaystyle 1^\infty$ is what is called an indeterminate form. It's value cannot be determined. In this problem, it turns out it is "equal" to e^2. In another problem, it might turn out to be "equal" to sin(1). Some other common indeterminate forms are $\displaystyle \frac{0}{0}$ and $\displaystyle \infty - \infty$. Here is a link to a Wolfram article on indeterminate forms: Indeterminate -- from Wolfram MathWorld