Showing $f_n(x):=\frac{x}{1+n^2x^2}$ uniformly convergent in $\mathbb R$ using $\epsilon-n_0$

Question

I want to show $$f_n(x):=\frac{x}{1+n^2x^2}$$ is uniformly convergent in $\mathbb R$ using $\epsilon-n_0$ argument.

So $|f_n-f|\leq\frac{|x|}{1+n^2|x|^2}$

I know $|f_n-f|\leq\frac1n$ since $\frac1n$ is the maximum of $f_n$. I can show this using the derivatives of $f_n$.

My question is: Can you show this or any other equation (independent of $x$) without using derivatives in any calculation?

I've tried $|f_n-f|\leq \frac1n \iff -\frac1n\leq\frac{x}{1+n^2x^2}\leq\frac1n \iff\ldots$ but I get no inequality which is obviously right.

Answer 1

Since

$$\frac x{1+n^2x^2}\xrightarrow[n\to\infty]{}0\;\;,\;\;\forall\,x\in\Bbb R$$

You want to prove that

$$\forall\,\epsilon>0\;\;\wedge\;\;\forall\,x\in\Bbb R\;\;\exists\,N_\epsilon\in\Bbb N\;\;s.t.\;\;n>N_\epsilon\implies\;\left|\frac x{1+n^2x^2}\right|<\epsilon$$

But

$$\left|\frac x{1+n^2x^2}\right|=\frac{|x|}{1+n^2x^2}<\epsilon\iff n^2>\frac1{|x|\epsilon}-\frac1{x^2}$$

Thus we can choose

$$\;N_\epsilon:=\frac1{2\epsilon}\;,\;\;\text{so}\;\;n>N_\epsilon\implies n^2>\frac1{4\epsilon^2}\ge\frac1{|x|\epsilon}-\frac1{x^2}\;$$

$$\text{Why? Check the maximum of the function}\;\;f(x)=\frac1{x\epsilon}-\frac1{x^2}$$

once for $\;x>0\;$ and once for $\;x<0\;$ .

Answer 2

You can show that $\frac x{1+n^2x^2}\leq \frac 1n$ by just multiplying with both denominators. Then, you have to show that $a\leq 1+a^2$ (with $a=xn$). Whit is true, because when $a\geq 1$, $a^2\geq a$, and when $a\leq 1$, $a\leq 1+a^2$. For the other inequality, you have to show that $a\geq -1-a^2$. Substitute $b=-a$ and multiply both sides with $-1$: $b\leq 1+b^2$. From above, we know this is true.

Answer 3

Yet a third approach could be to find the maximum value of $f_{n}$ for each $n$. obviously the limit as $\vert x \vert \to \infty$ of $f_{n}(x)$ is zero. Furthermore, taking the derivative gives $$ f^{\prime}_{n}(x)= \frac{1-n^2x^2}{(1+n^2x^2)^2}$$ The denominator is never zero and the numerator is zero precisely when $x=\pm \frac{1}{n}$. One easily finds $\frac{1}{n}$ to be the critical point at which $f_{n}$ achieves a max. The value of $f_{n}$ at this value is $\frac{1}{2n}$. Since this is a global max, it bounds $f_{n}$ uniformly from above. The minimum occurs at the other critical point which gives you a bound below.