lim$_{n\rightarrow \infty}\int _{-\pi}^\pi f(t)\cos nt\,dt$ let $f\in C[-\pi,\pi]$
i.e $f$ is continuous over $[-\pi,\pi]$
Evaluate:lim$_{n\rightarrow \infty}\int _{-\pi}^\pi f(t)\cos(nt)\,\mathrm dt$
I tried to evaluate it using normal integration technique but that yielded nothing.Any hints
Another one is lim $_{h\rightarrow 0}\frac{1}{h}\int _{-h}^hf(t)\,dt $  How to do this?
 A: Trying to bring back old memories of my teacher's very good elementary proof ... I'll try to give a sketch and hopefully some intuition for 1. (2. is trivial as per @Jason's comment):
First note that your integral can be written as 
$$\int _{-\pi}^{\pi} f(t)\cos(nt)dt = 
\frac{1}{n} \sum_{k=-n}^{n-1} \int_{k\pi}^{(k+1)\pi} f(u/n)\cos(u)du = \\ 
= \frac{1}{n} \sum_k (-1)^k \int_0^\pi f\left(\frac{v+k\pi}{n}\right)\cos(v)dv \tag{1}$$
Now when $n \rightarrow \infty$, $v \mapsto f\left(\frac{v+k\pi}{n}\right)$ becomes very "flat" over $[0,\pi]$ while over the same interval $\int_0^{\pi} \cos(v)dv = 0$. So intuitively you would like to see something like 
$$\int_0^\pi f\left(\frac{v+k\pi}{n}\right)\cos(v)dv \sim \int_0^\pi f\left(\frac{k\pi}{n}\right)\cos(v)dv =0$$ 
To get to that, take $\epsilon > 0$. Since $f$ is continuous on $[0,\pi]$, it is also uniformly continuous (from the Heine-Cantor theorem), so there is a $\eta = \eta_{\epsilon} > 0$ so that $\lvert x-y \rvert < \eta \implies \lvert f(x)-f(y) \rvert < \epsilon$. 
Now let's take an integer $n > N_{\epsilon} = \tfrac{\pi}{\epsilon}$. You would have for all $k$'s: 
$$\forall v \in [0,\pi], \left| f\left(\frac{v+k\pi}{n}\right) - f\left(\frac{k\pi}{n}\right)\right| < \epsilon$$
and thus (oh what a mouthful of LaTeX ...):
$$ \left| \int_0^\pi f\left(\frac{v+k\pi}{n}\right)\cos(v)dv \right| =
\left| \int_0^\pi f\left(\frac{v+k\pi}{n}\right)\cos(v)dv -
\int_0^\pi f\left(\frac{k\pi}{n}\right)\cos(v)dv \right| \le \\ 
\int_0^\pi \left| f\left(\frac{v+k\pi}{n}\right) -
f\left(\frac{k\pi}{n}\right)\right| |\cos(v)|dv \le
\epsilon \pi $$
Now inject that in $(1)$ and you get:
$$ n\ge N_{\epsilon} \implies \left|\int _{-\pi}^{\pi} f(t)\cos(nt)dt \right| \le 2\pi\epsilon$$
And you conclude $$\lim_{n\rightarrow\infty} \int _{-\pi}^{\pi} f(t)\cos(nt)dt = 0$$
Now I hope I haven't ridiculed myself with some major mistake... 
The good thing with this proof is that you actually see what topological and functional properties are required to get the result:


*

*the interval $[0,\pi]$ is compact in $\mathbb R$

*$f$ is continuous and thus uniformly continuous

*$\cos$ verifies $\int \phi = 0$ and $\int |\phi| < \pi$ on each interval $[k\pi,(k+1)\pi]$


It's probably easy to find examples that show that relaxing any of these conditions would void the result. 
A: Nice one Alex H. :) , I have a proof (that is, in terms of topological tools, close to yours) using a different implementation.
I use two results:
-Firstly, if the hypothesis on the function is $f\in C^1[-\pi,\pi]$ then you get the result with an integration by part on the cos, and deriving the function. Very easy to prove.
-Secondly, I'll use the Weirstrass theorem: there is a sequence of polynomials that uniformly converges $\rightarrow f$. I'll note the sequence $P_k$.
Now: $| \int_{-\pi}^{\pi} f(t)\cos(nt)dt - \int_{-\pi}^{\pi} P_k(t)\cos(nt)dt | = |\int_{-\pi}^{\pi} (f(t)-P_k(t))\cos(nt)dt| \leq 2\pi |f-P_k|_{\infty}$ 
So, for a certain N positive integer, we get:
$ k \geq N: | \int_{-\pi}^{\pi} f(t)\cos(nt)dt - \int_{-\pi}^{\pi} P_k(t)\cos(nt)dt | \leq \epsilon $ ;  thanks to the uniform convergence of the ($P_k$)
From the first point we know that : $ \int_{-\pi}^{\pi} P_k(t)\cos(nt)dt \rightarrow 0 $ , when $n \rightarrow +\infty$ , since every polynomial is $C^1[-\pi,\pi]$.
Using | |a| - |b| | $\leq |a-b|$ we get : $|\int_{-\pi}^{\pi} f(t)\cos(nt)dt| \leq \epsilon + |\int_{-\pi}^{\pi} P_k(t)\cos(nt)dt| \leq 2\epsilon $ , for k>N, n > $N_1$ set with the above result. 
Hence you get what you wanted 
