Check Uniform convergence of $f_{n}\left(x\right)=\ln\left(1+\frac{1}{nx^2}\right)$ on $(0,1]$

Question

Check Uniform convergence of $f_{n}\left(x\right)=\ln\left(1+\frac{1}{nx^2}\right)$ on $(0,1]$.

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Attempt:

$$f(x)=\lim _{n\to \infty}\ln\left(1+\frac{1}{nx^2}\right)=\ln(1)=0$$

now I need to find if :

$$\lim _{n\to \infty}\sup_{x\in (0,1]}|f_{n}(x)-f(x)|=0$$

and here I stuck I don't know if $f(x)$ that I found is correct and how to continue from here ?

I thought of $g(x)=\ln(1+\frac{1}{nx^2})$ so $g'(x)=0$... but I don't think this will help.

Thanks.

Answer 1

the convergence cannot be uniform $$\lim _{n\to \infty }\sup_{x\in (0,1]}|f_{n}(x)|\overset{x=1/n}{\ge} \lim _{n\to \infty } f(1/n)=\lim _{n\to \infty }\ln(1+n))=\infty$$

or alternatively $$\lim _{n\to \infty }\sup_{x\in (0,1]}|f_{n}(x)|\overset{x=1/\sqrt{n}}{\ge} \lim _{n\to \infty } f(1/\sqrt{n}) =\ln 2$$

but rather the convergence is uniform on any other compact subset of $(0,1]$ since $$\sup_{[a,1]}|f_n(x)|\le |f_n(a)|\to0,~~a>0$$

Answer 2

It cannot be uniform convergent. Take $x_n=\frac{1} {n}\in(0,1] $, but we have $$f_n(x_n) =\ln\left( 1+n\right) \to \infty$$ as $n\to\infty$. Hence $f_n$ not uniform convergent.