A limit of an indexed integral at closed interval

[BaitiTamam]

Let ρ(x) be a continuous function on ℝ, which evaluates ρ(x)=0 when |x|≥1 and that meets following.

∫[-1,1]ρ(x)dx=1

And let ψ(x) be a continuous function on interval [-1,1], prove

lim[n→∞] n∫[-1,1]ρ(nx)ψ(x)dx = ψ(0).

is denoted.


This is NOT a homework but a past exam problem of a college that has no answer.
I've been thinking for a whole day now.

What I have found so far:

First I defined sequencing functions as following:

F_n(x)=∫[-1,x]ρ(ny)ψ(y)dy
G_n(x)=∫[x,1]ρ(ny)ψ(y)dy

Now given a properly large natural number of n_0 which completes n_0≥1/y (where y≠0),
for all n∈N of n≥n_0, ρ(ny) evaluates 0. In this sense, taken a small positive number ε>0 instead of y, we only have to concern that of a small closed interval of [-ε,ε] as to nF_n(-ε) and nG_n(ε) gives 0 (for all n≥n_0≥1/ε).

So for all n same above,

n∫[-1,1]ρ(nx)ψ(x)dx = n∫[-ε,ε]ρ(nx)ψ(x)dx


And I have a stuck.
Please help ;)

 

[lurflurf]

First change variables (just for simplicity)
n∫[-1,1]ρ(nx)ψ(x)dx =∫[-1,1]ρ(x)ψ(x/n)dx
now by the mean value theorem for integrals
∫[-1,1]ρ(x)ψ(x/n)dx =ψ(t/n)∫[-1,1]ρ(x)dx =ψ(t/n)
where |t|<1, but may depend upon ψ and n
consider
lim[n→∞] ψ(t/n)

 

[BaitiTamam]

>lurflurf

I got it. Changing variable was out of my thought.
Thank you:)