$f_n(x) = \log(1 + n^3 x^2)/n^2$ (uniform converge)

Question

$$ \mbox{Let}\quad \operatorname{f}_{n}\left(x\right) = \frac{\log\left(1 + n^{3}x^{2}\right)}{n^{2}} $$ Show that $\left(\operatorname{f}_{n}\right)_{n}$ converges uniformly to the zero function on $\left[0,1\right]$.

I know that uniform convergence mean

 $$ \left\vert\operatorname{f}_{n}\left(x\right)−\operatorname{f}\left(x\right)\right\vert < \varepsilon, \qquad \forall n \geq N,\ \forall x\in \left[0,1\right].$$

Can someone help me prove it.

Answer 1

For $ n>0$,

$$x\mapsto 1+n^3x^2 $$ is increasing from $ [0,1] $ to $[1,+\infty)$

$$x \mapsto \ln(x) $$ is increasing at $ [1,+\infty)$,

So, $ f_n $ is increasing at $ [0,1]$ and

$$M_n=\sup_{x\in [0,1]}|f_n(x)|=f_n(1)$$ $$=\frac{\ln(1+n^3)}{n^2}$$ with

$$0\le M_n\le \frac{\ln(n^3+n^3)}{n}=\frac{\ln(2)+3\ln(n)}{n}$$ and by squeezing,

$$\lim_{n\to +\infty}M_n=0$$

the convergence to zero is uniform at $ [0,1]$.

Answer 2

Hint: $\sup\limits_{x\in [0,1]}|f_n(x)-0|\rightarrow 0$ as $n\rightarrow\infty.$