Check: $\lim_{n\to \infty} f^{'}_n(x)=f^{'}(x)$

Question

I've got $f_n(x)=\frac{\sqrt{(1-x^2)^{n^2+1}}}{n} , \forall{x}\in{[-1,1], \forall{n}\in{\mathbb{N}}}$

I've analyzed that $f_n(x)$ is pointwise and uniformly convergence in $[−1,1]$.

Now, I have to check that $\lim_{n\to \infty} f^{'}_n(x)=f^{'}(x)$ being $f(x)=0$ ( I have already calculated it) the point limit of $f_n(x)$ in $[−1,1]$.

---

My doubt is if to check it I need to see the 3 conditions of the theorem of uniform convergence and differentiable or it's enough to calculate the limit and check that it is 0. In this second case, how can I calculate the limit?

$\lim_{n \to \infty} f^´_n(x)=\frac{-(n^2+1)x(1-x^2)^{\frac{n^2-1}{2}}}{n}$

Answer 1

For a fixed $x \in [-1, 1]$, you can calculate the limit as follows:

1. $x = 0.$ In this case, you have $f_n'(0) = 0$ for all $n \in \Bbb N$ and thus, the limit does agree.

2. $x \neq 0.$ In this case, you have $1 - x^2 < 1$ and thus, $(1 - x^2)^{(n^2+1)/2}$ goes to $0$ rapidly enough so that

$$\lim_{n \to \infty}\frac{-(n^2+1)x(1-x^2)^{\frac{n^2-1}{2}}}{n} = 0.$$

(This follows essentially from the fact that $n^2\chi^n \to 0$ as $n\to\infty$ for any $|\chi| < 1$.)