Why are these limits finite? I am solving a test and have trouble solving this particular task: given a differentiable function $f : \mathbb{R} \rightarrow \mathbb{R} $ and $ f(0) = f(1) = 0$, are the following limits finite:
$$
(a) \lim_{n \rightarrow +\infty} nf\left(\frac{1}{n}\right)
$$
$$
(b) \lim_{n \rightarrow +\infty} nf\left(\frac{n+1}{n}\right)
$$
$$
(c) \lim_{n \rightarrow +\infty} nf\left(\frac{n}{n+1}\right)
$$
Why are those limits finite (I know they all are)?
 A: The following is rendered useless after the OP added the information that $f$ is differentiable However, I leave the answer, as it may show (at least to the OP, who didn't bother to mention it), that without differentiability the answer may be entirely different.
You have not specified whether $f$ is continuous. If it is not, those limits need not even exist, and if they do, they may be anything.
Let's assume $f$ is continuous. Then, we know that $f(0)=f(1)=0$, and, since $f$ is continuous, $f(x)$ has to be small around $0$ and $1$. How small? Not enough, I fear.
For instance, let $f(x)=\sqrt{x(1-x)}$, then as $n\to\infty$, we have
$$nf(1/n)=n\sqrt{\frac{1}{n}\left(1-\frac{1}{n}\right)}=\sqrt{n}\sqrt{1-\frac{1}{n}}\to\infty$$
Likewise, $f(n/(1+n))\to\infty$. It would be easy to find an example for which the third limit is infinite as well. It's also easy to make up an $f$ such that these limits do not exist: for instance, make $f(1/n)$ of the order of $1/\sqrt{n}$ as $n\to\infty$, and oscillating.
A: Hint: use $f(x+h)=f(x)+f'(x)h+o(h)$ when $h\to 0$.
The crucial fact is that $f$ is differentiable in $0$ and $1$, that by definition mean it can be approximated by linear function at this points.
A: a) 
$$
\lim_{n\to\infty}n\,f\Bigl(\frac1n\Bigr)=\lim_{n\to\infty}\frac{f(1/n)-f(0)}{1/n}=f'(0).
$$
This can be adapted to treat b) and c).
