Show that function is continuous at $x_0$ I want to show that $f: \mathbb R \rightarrow \mathbb R$,
$f(x) = \frac{x-1}{x^2+1}$ is continuous at $x_0=-1$. So if $|x+1|<\delta$, then $|f(x)+1| < \epsilon$.
I rearranged $|f(x)+1|=|\frac{(x-1) +(x^2+1)}{x^2+1}|=|\frac{x+x^2}{x^2+1}|=|\frac{1+x}{x+ \frac{1}{x}}|$, but now I'm stuck. Can anybody help me? 
Thanks in advance!

You can use that $\displaystyle\left|\frac{x+x^2}{x^2+1}\right|=\frac{|x||x+1|}{x^2+1}<|x+1|$ for all x, since $|x|<x^2+1$ 
because $x^2<(x^2+1)^2=x^4+2x^2+1$ for all x.
