Finding limit of a continuous function I have difficulty in calculating the following limit.
Let $f$ be a continuous function on $[-1,1]$, it is desired to find the following limit,
$\lim_{n\rightarrow \infty}n \int_{-\frac{1}{n}}^{\frac{1}{n}} f(x)(1-n|x|)dx$
Thank you very much in advance.
 A: Firstly,
$$
\lim_{n\rightarrow \infty}n \int_{-\frac{1}{n}}^{\frac{1}{n}} f(x)(1-n|x|)dx = \\
\lim_{n\rightarrow \infty}nf(\xi_n)\cdot\int_{-\frac{1}{n}}^{\frac{1}{n}}(1-n|x|)dx
$$
by the intermediate value theorem, for $\xi_n \in (-\frac{1}{n},\frac{1}{n})$. Now, since
$$
\int_{-\frac{1}{n}}^{\frac{1}{n}}(1-n|x|)dx = 2\int_{0}^{\frac{1}{n}}(1-nx)dx = 2(\frac{1}{n} -n\frac{1}{2n^2}) = \frac{1}{n}
$$
we have 
$$
\lim_{n\rightarrow \infty}n \int_{-\frac{1}{n}}^{\frac{1}{n}} f(x)(1-n|x|)dx = \lim_{n\rightarrow \infty}f(\xi_n) = f(0)
$$
where the last equality holds because since $f$ is continuous and $\xi_n \to 0$, $f(\xi_n) \to 0$.
A: The limit, call it L, is $f(0)$.  To see this, fix $\epsilon$.  Then, by continuity of $f$, $\exists N \in \mathbb{N}$ such that $$ n > N \Rightarrow f(0) - \epsilon \le f(x) \le f(0)+ \epsilon $$ on the interval $I_n= [\frac{-1}{n},\frac{1}{n}] $.
Now, multiplying with $n(1-n|x|)$ and integrating over the interval $I_n= [\frac{-1}{n},\frac{1}{n}] $ we get that for all $n > N$,
$$(f(0)-\epsilon) \int_{I_n}   n(1-n|x|) \le     \int_{I_n}  f(x) n(1-n|x|) \le  (f(0)+\epsilon)\int_{I_n}  n(1-n|x|)$$ The integrals on each side are trivial to evaluate (hint: What’s the area of a triangle of height $n$ and base $\frac{2}{n}$?).
$$f(0)-\epsilon \le      n  \int_{I_n} f(x) (1-n|x|)  \le  f(0)+\epsilon  $$
Now, taking the limit,
$$ f(0)-\epsilon \le L \le  f(0)+\epsilon$$ 
Since $\epsilon$ is arbitrary, we have $$L = f(0)$$
A couple of related notes for intuition:  


*

*In kernel smoothing of functions, your function $n(1-n|x|)$ is called the triangle kernel or the linear kernel.  As the kernel bandwidth goes to zero, convolution with it must reproduce the function value $f(0)$. 

*Also, more loosely speaking, the limit as $n \rightarrow \infty$ of your kernel is the Dirac delta "function". Convolution by it recovers the function value at 0. 
