Limits for sequences and the $e$ method After doing a lot of homework, I've realized that I don't really get how to use the $\left(1+\frac{1}{n}\right)^n\to e$ method, which shows in how I can't solve indeterminate limits of the type $1^\infty$. For example:
$$\lim_{n\to\infty}{\left(\frac{1}{2}+\frac{2}{n}\right)^n}$$
Then I would do as follows:
$$\lim_{n\to\infty}{\left(\frac{1}{2}+\frac{1}{2}+\frac{2}{n}-\frac{1}{2}\right)^n}$$
$$\lim_{n\to\infty}{\left(1+\frac{4-n}{2n}\right)^n}$$
$$\lim_{n\to\infty}{\left(1+\frac{1}{\frac{2n}{4-n}}\right)^n}$$
And, from here on, I would not know what to do. Thus, the question is:
How am I supposed to work with the exponents?
 A: Write $$\left(\frac{1}{2} + \frac{2}{n}\right)^n = \frac{1}{2^n} \left(1+\frac{4}{n}\right)^n$$
Taking limits, we get
$$\lim_{n \rightarrow \infty} \left(\frac{1}{2} + \frac{2}{n} \right)^n = 0 \cdot e^4 = 0$$
A: Your final limit, $$\lim_{n\to\infty}{\left(1+\frac{1}{\frac{2n}{4-n}}\right)^n}\;,$$ does not have the right form for you to apply the fact that $\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$, because the fraction $$\frac{1}{\frac{2n}{4-n}}$$ does not tend to $0$ as $n\to\infty$. Neither does the fraction $\dfrac{4-n}{2n}$ from the preceding step. Thus, either you need a different rearrangement altogether, or you need to use a different approach. As others have already suggested, a different approach does the job nicely.
A: Another approach:
If $\lim_{x\to{+\infty}} f(x)^{g(x)}$ is as $costant^{+\infty}$ which is indeterminate form then: $$\lim_{x\to{+\infty}} f(x)^{g(x)}=e^{\lim_{x\to +\infty}\big(f(x)-1\big)g(x)}$$
A: Here's a trick:
$$\lim_{n\rightarrow \infty}\left(\frac{1}{2}+\frac{2}{n}\right)^n =\lim_{n\rightarrow \infty}e^{n\ln\left(\frac{1}{2}+\frac{2}{n}\right)}$$ which is $\,0\,$ since the limit of $$n\ln\left(\frac{1}{2} + \frac{2}{n}\right)$$ as n tends to infinity is $-\infty$. 
Note, you can also use the trick above to prove that 
$$\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n = e$$
