# Why are these limits finite? [closed]

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)?

## closed as off-topic by Jean-Claude Arbaut, Parcly Taxel, John B, user21820, Daniel W. FarlowSep 18 '16 at 13:55

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• MathJax tip: You can use \left( "expression in parenthesis" \right) to make nicer parentheses. – Bobson Dugnutt Sep 18 '16 at 9:02
• Is $f$ continuous? – Canis Lupus Sep 18 '16 at 9:06
• @Corvus It's differentiable. – syntagma Sep 18 '16 at 9:07
• That is a crucial piece of information which you should have put in your question. – Hirshy Sep 18 '16 at 9:08
• I know, I've edited the question. – syntagma Sep 18 '16 at 9:09

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).

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.

• The limits also need not exist if $f$ is continuous. – Carsten S Sep 18 '16 at 10:37
• @CarstenS Maybe I was not clear enough, but it's what I wrote ;-) – Jean-Claude Arbaut Sep 18 '16 at 17:18
• I have actually only read: You have not specified that whether $f$ is continuous. If it is not, those limits need not even exist, and if they do, they may be anything. I still do not understand what you are trying to say there. – Carsten S Sep 18 '16 at 22:19
• @CarstenS If $f$ is not continuous, the value $f(0)$ tells nothing about the values around $0$, let alone its limits. If however, $f$ is continuous, it has a limit everywhere, and if $f(0)=0$, at least you know $f$ has to be small around $0$, so you may attempt to prove something, but as the square root shows, $f$ is not necessarily small enough. I am just repeating what I have written above. – Jean-Claude Arbaut Sep 19 '16 at 5:55

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.

• Not only can it be approximated by a linear function, it is (locally) linear. – bjb568 Sep 18 '16 at 13:31
• @bjb568: strictly speaking, no -- you can't simply ignore $o(h)$ in general case. – Canis Lupus Sep 18 '16 at 13:52
• Of course, I'm just saying it's a line, not just close to a line, if you're looking at it closely enough. – bjb568 Sep 18 '16 at 17:33