How does this inequality help me evaluate the limit? Suppose $b>0$. Compute 
$$
\lim_{n \rightarrow \infty} n\left(\left(\frac{n+1}{n}\right)^b-1\right). 
$$
My professor said that $e^{\frac{1}{n+1}} \leq 1 + 1/n$ can be used to evaluate the limit, but I do not see how this is done. 
Now $\lim_{n \rightarrow \infty} n((e^{\frac{1}{n+1}})^b-1) \leq \lim_{n \rightarrow \infty} n((\frac{n+1}{n})^b-1)$, but I believe $n((e^{\frac{1}{n+1}})^b-1)$ converges, (Right?).
A hint as to how one should proceed would be appreciated. Thanks. 
 A: Show that $e^{\frac{1}{n+1}} \leq 1 + \frac1n \leq e^\frac1n$. Then you can use the squeeze theorem to solve the problem.
A: Do you know approximations? because one can treat this question using:
$(1+\frac{1}{n})^b = 1 + \frac{b}{n} + o(\frac{1}{n})$
This gives you : $ n*[(1+\frac{1}{n})^b -1] =  b + o(1) \rightarrow b $ directly
As for your inequality,, I'm just not sure that it's the best approach. At least combine it with:
$\frac{1}{n}-\frac{1}{2n^2} \leq ln(1+\frac{1}{n}) \leq \frac{1}{n} $.
You can get your limit this way I think, if you note that: $ (\frac{n+1}{n})^b = e^{b*ln(1+\frac{1}{n})} $
A: In other way, if $k=\lceil b\rceil$ then $b\leq k$, then 
$$\lim_{n \rightarrow \infty} n\left[\left(\frac{n+1}{n}\right)^b-1\right]\leq \lim_{n \rightarrow \infty} n\left[\left(\frac{n+1}{n}\right)^k-1\right]$$
How
$$\lim_{n \rightarrow \infty} n\left[\left(\frac{n+1}{n}\right)^k-1\right]=\lim_{n \rightarrow \infty} n\left[\frac{n^k+\binom{k}{1}n^{k-1}+\cdots+1}{n^k}-1\right]=\lim_{n \rightarrow \infty} n\left[\frac{n^k+kn^{k-1}+\cdots+1-n^k}{n^k}\right]=\lim_{n \rightarrow \infty} \left[\frac{kn^{k}+\cdots+n}{n^k}\right]=k$$
then: 
$$\lim_{n \rightarrow \infty} n\left[\left(\frac{n+1}{n}\right)^b-1\right]$$ converges.
A: By L'Hopital, we have:
$$\lim_{x\rightarrow 0}\frac{(1+x)^b-1}{x}=\lim_{x\rightarrow0}b(1+x)^{b-1}=b.$$
Take $x_n=\frac{1}{n}$, and you get:
$$\lim_{n\rightarrow\infty}n\left[(\frac{n+1}{n})^b-1\right]=b.$$
