How can I show $f_{n}(x) = \sqrt{x^{2} + \frac{1}{n}}$ is continuously differentiable on $x \in ]-1,+1[$ for each $n \in \mathbb{N}$?

Question

I want to show $f_{n}(x) = \sqrt{x^{2} + \frac{1}{n}}$ is continuously differentiable for each $x \in ]-1,+1[$ and $n \in \mathbb{N}$?

I have

$$f'_{n}(x) = \frac{x}{\sqrt{x^{2} + \frac{1}{n}}}.$$

It is obviously continuous because the denominator is always positive. But is there a more rigorous way to show this? Or is this sufficient? Since $n \in \mathbb{N}$, the $ \frac{1}{n}$ is always positive. Also $x^{2}$ is the square of a number, so it's always positive. I just want to show continuity though.

Answer 1

I believe that's enough.

It should be clear that $x^2+\frac1n$, a polynomial, is continuous. Composition of continuos function is continuous, Hence $\sqrt{x^2+\frac1n}$ is continuous.

A continuous function divided by a non-zero continuous function gives us another continuous function, hence your function is continuous.