Sequence of function which converges pointwise, is it also uniform?

Question

The question is:

Does the sequence $$f_k \left( x \right) = \dfrac{1}{k(1 + x^2)}$$ converge pointwise on $\Bbb{R}$?Is the convergence uniform?

For the first part of the question I done $\lim\limits_{k \rightarrow \infty} \dfrac{1}{k(1 + x^2)} = 0$, so it converges pointwise on the function $$f \left( x \right) = 0. $$ I am not too sure on how do the second part, which is to see if it converges uniformly or not. Any help would be appreciated.

Answer 1

For any $k\in\Bbb N$ we have $(1+x^2)\ge 1\land (1+x^2)>0\Rightarrow0<\dfrac{1}{1+x^2}\le1\Rightarrow0<\dfrac{1}{k(1+x^2)}\le\dfrac{1}{k}\Rightarrow\sup_{x\in\Bbb R}f_k(x)\le\dfrac{1}{k}.$

Since $\dfrac{1}{k(1+x^2)}=\dfrac{1}{k}$ for $x=0$, then $\sup_{x\in\Bbb R}f_k(x)=\dfrac{1}{k}.$

Then $|f_k(x)|\le \sup_{x\in\Bbb R}|f_k(x)|=\sup_{x\in\Bbb R}f_k(x)=\dfrac{1}{k}$, so $\lim_{k\to\infty}|f_k(x)-f(x)|=\lim_{k\to\infty}|f_k(x)|\le\lim_{k\to\infty}\dfrac{1}{k}=0$.

Answer 2

For each $k\in\Bbb N$, $\sup_{x\in\Bbb R}\dfrac1{k(1+x^2)}=\dfrac1k$. Since $\lim_{k\to\infty}\frac1k=0$, the convergence is uniform.