# Convergence and Differentiability of a Sequence of Functions

Question:

I'm studying a sequence of functions $$f_n(x) = \sqrt{x^2 + \frac{1}{n}}$$ defined on the domain $$[-1,1]$$ and I'm trying to understand their behavior as $$n$$ approaches infinity.

Context and Motivation:

This problem came up while I was studying real analysis and the concept of uniform convergence. I understand that for a sequence of functions to converge uniformly to a limiting function, the speed of convergence must be independent of the choice of $$x$$ in the domain.

I'm interested in seeing how this plays out for the given sequence of functions, especially since they involve a square root, which can often lead to interesting behavior.

My Attempts:

I've tried to find the pointwise limit of the sequence by taking the limit as $$n$$ goes to infinity, which gives me the function $$f(x) = |x|$$.

However, I'm unsure if the convergence is uniform. I've also noticed that each $$f_n$$ is differentiable, but I'm not sure if the limiting function $$f$$ is differentiable on $$[-1,1]$$ due to the absolute value.

Specific Questions:

1. Does the sequence $$f_n(x)$$ converge uniformly to $$f(x)$$ on $$[-1,1]$$?
2. Is the limiting function $$f(x)$$ differentiable on $$[-1,1]$$?

For uniform convergence, notice that $$f(x)=|x|=\sqrt{x^2}$$.
Thus, $$f_n(x)-f(x)=\sqrt{x^2+1/n}-\sqrt{x^2}=\dfrac{1/n}{\sqrt{x^2+1/n}+\sqrt{x^2}}\leq \dfrac{1/n}{\sqrt{1/n}}=\dfrac{1}{\sqrt{n}}$$
Therefore, we can bound the difference with no dependence on the point $$x$$, and we have uniform convergence.
But $$f(x)=|x|$$ is not a differentiable function on $$x=0$$, which is a standard result.