Given is the sequence of functions $f_n(x)= \sqrt{\frac1{\sqrt n} + x^2}$ with x element of the real numbers.
If I am correct, the following is the limit value $$f(x) = \lim f_n(x) = \lim \sqrt{\frac1{\sqrt n} + x^2} = |x|$$
Unfortunately, it is not clear to me how I can now show the uniform convergence.
The beginn would be: $$|f_n(x) -f(x)| = |\sqrt{\frac1{\sqrt n} + x^2} - |x||$$
How does the assessment go on?
Your limit for the pointwise convergence is not correct: $$\lim\limits_{n\rightarrow\infty}f_n(x)=\lim\limits_{n\rightarrow\infty}\sqrt{\frac{1}{\sqrt{n}}+x^2}=\sqrt{\lim\limits_{n\rightarrow\infty}\frac{1}{\sqrt{n}}+x^2}=\sqrt{x^2}=|x|$$ since $y\mapsto\sqrt{y}$ is continuous on $[0,\infty)$
So the function is continuous on $\mathbb R$ and thus the sequence has a candidate for a uniform limit.
Hint: Regarding the uniform convergence, we aim to show/or disprove that $\lim\limits_{n\rightarrow\infty}\sup_{x\in\mathbb{R}}|f_n(x)-f(x)|=0$. Observe the difference: $$|f_n(x)-f(x)|=\left| \sqrt{\frac{1}{\sqrt{n}}+x^2}-|x|\right|$$ on different intervals at first (they are intervals where it is differentiable...), what can you say?