# Thread: Given certain conditions, find limit?

1. ## Given certain conditions, find limit?

If $f : [-1, 1] \rightarrow \mathbb{R}$ and $f'(0) = \lim_{n\rightarrow \infty} nf\left(\frac{1}{n}\right)$ and $f(0) = 0$, find the value of $\lim_{n\rightarrow \infty} \left\{\frac{2}{\pi}(n + 1)\arccos \left(\frac{1}{n}\right) - n\right\}$.

2. ## Poorly worded question

Define $f(x)$ such that $nf(\frac1n)=\frac2\pi(n+1)\arccos(\frac1n)-n$, so $f(x)=\frac2\pi(x+1)\arccos x -1$ - checking, $f(0)=0$, as required. So $f'(x)=\frac2\pi\left[\arccos x-(x+1)\frac1{\sqrt{1-x^2}}\right]$, making $f'(0)=1-\frac2\pi$

3. Originally Posted by fardeen_gen
If $f : [-1, 1] \rightarrow \mathbb{R}$ and $f'(0) = \lim_{n\rightarrow \infty} nf\left(\frac{1}{n}\right)$ and $f(0) = 0$, find the value of $\lim_{n\rightarrow \infty} \left\{\frac{2}{\pi}(n + 1)\arccos \left(\frac{1}{n}\right) - n\right\}$.
What has the "if" conditions got to do with the limit?

There is no f(x) in the limit!

4. ## Theorem?

I think the problem should read: If $f : [-1, 1] \rightarrow \mathbb{R}$ and $f(0) = 0$, then $f'(0) = \lim_{n\rightarrow \infty} nf\left(\frac{1}{n}\right)$ . Use this theorem to evaluate the following limit...

This "theorem" makes sense because $f'(0)=\lim_{h\to0}\frac{f(0+h)-f(0)}{h-0}=\lim_{h\to0}\frac1h f(h)=\lim_{n\to\infty}nf(\frac1n)$

It was just a poorly worded question, but an interesting technique I had not seen before.