Proof continuity of a function with epsilon-delta I quickly need help with a problem that seems to be fairly easy but I can't really do the final step:
Proof that the function $\frac{x-1}{x²+1}$ is continuus in $x = -1$
using the epsilon-delta-definition.
Thanks in advance.
 A: HINT:
First write 
$$\left|\frac{x-1}{x^2+1}+1\right|=\left|\frac{x(x+1)}{x^2+1}\right|\le |x|\,|x+1|$$
Second, restrict $x$, say for example $-3/2\le  x\le -1/2$.
Can you finish now?
A: Your delta should be picked carefully so that $\dfrac{x}{x^2+1}
$ is nicely bounded.  
A: Let $\epsilon >0$. We want to pick a $\delta$ that such that
$$
\bigg|\frac{x-1}{x^{2}+1}+1\bigg|=\bigg|\frac{x(x+1)}{x^{2}+1}\bigg|\leq|x||x+1|<\epsilon.
$$ 
We are free top choose $\delta$ as we wish, and, in doing so, we will have a handle on making the $|x+1|$ term above as small as we want. To help us minimize $|x|$, let's see what happens if we would ensure $|x+1|<1/2$.
In this case, 
$$
|x+1|<\frac{1}{2}\implies-\frac{1}{2}<x+1<\frac{1}{2}\implies-\frac{3}{2}<x<-\frac{1}{2}<\frac{3}{2}\implies |x|<\frac{3}{2}.
$$
Thus, if $|x+1|<1/2$, it follows that $|x||x+1|<\frac{3}{2}|x+1|$. If, in addition, we ensure that $\delta<\frac{2\epsilon}{3}$, it follows that $|x||x+1|<\frac{3}{2}\cdot\frac{2\epsilon}{3}=\epsilon$.
Therefore, choosing $\delta=\min\{\frac{1}{2},\frac{2}{3}\epsilon\}$ works. 
