<s>[INST] Showing that the function $f(x) = \frac{x}{1+x^2}$ is continuous using the $\epsilon \delta$ definition I am attempting to prove that $f(x) = \frac{x}{1+x^2}$ is continuous. Here's my attempt so far.
In order to prove that $f(x)$ is continuous, it is necessary to show that for every $\epsilon>0$ there exists a $\delta>0$ such that $|\frac{x}{1+x^2}-\frac{c}{1+c^2}|<\epsilon$ when $|x-c|<\delta$.
I first start by working with the left side of the inequality. We can say that $$|\frac{x}{1+x^2}-\frac{c}{1+c^2}| = \frac{|x-c+c^2x-cx^2|}{|(1+x^2)(1+c^2)|}$$ After this we can use the triangle inequality to say, $$\frac{|x-c+c^2x-cx^2|}{|1+x^2||1+c^2|}\leq \frac{|x-c|+|c^2x|+|-cx^2|}{|1+x^2||1+c^2|}$$
Now we can break the function into 3 terms $$\frac{|x-c|}{|1+x^2||1+c^2|}+\frac{|c^2x|}{|1+x^2||1+c^2|}+\frac{|x^2c|}{|1+x^2||1+c^2|}$$
Then we use the fact that $$\frac{c^2x}{|1+x^2||1+c^2|}<1$$ which can be applied to the last 2 terms using the same logic. This gives $$\frac{|x-c|}{|1+x^2||1+c^2|}+2$$
Now we want to isloate $|x-c|$ in order to solve for $\delta$.
I have
$$\frac{|x-c|}{|1+x^2||1+c^2|}<\epsilon-2$$. This is where I get stuck. Any help would be appreciated, thanks!
 [/INST] When $c \neq 0$, use the bound
$$\tag{*}\left|\frac{x}{1+x^2}-\frac{c}{1+c^2}\right| = \frac{|x-c+c^2x-cx^2|}{(1+x^2)(1+c^2)} =\frac{|x-c+c^2x-cx^2+c^3 - c^3|}{(1+x^2)(1+c^2)} \\ \leqslant\frac{|x-c|+c^2|x-c|+|c||x^2-c^2|}{(1+x^2)(1+c^2)}= \frac{1+c^2+|c||x+c|}{(1+x^2)(1+c^2)}|x-c|$$
If $|x-c| < \frac{|c|}{2}$, then, since $|x| - |c| \leqslant |x-c|$, we have $\frac{|c|}{2} < |x| < \frac{3|c|}{2}$ and
$$\tag{**}|x+c| \leqslant |x|+|c| <\frac{5|c|}{2}, \quad 1+x^2 > 1+ \frac{c^2}{2}$$
Bounding on the RHS of (*) with (**), we get
$$\left|\frac{x}{1+x^2}-\frac{c}{1+c^2}\right| < \frac{1+c^2 + \frac{5c^2}{2}}{\left(1 + \frac{c^2}{2}\right)(1+c^2)}|x-c|$$
You should be able to finish from here by finding $\delta > 0$ such that if $|x-c| < \min(\delta, \frac{|c|^2}{2})$, then $|f(x) - f(c)| < \epsilon$.
Handle the case where $c=0$ separately.
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