Prove $\lim_{x\to 3^+}\frac{x^2+1}{x-3}=+\infty $ using definition Let $f(x)=\frac{x^2+1}{x-3}$. I have to prove that $$\forall M>0, \exists \delta>0: |x-3|<\delta\implies  f(x)>M$$
I proved that $$\frac{x^2+1}{x-3}=(x-3)+6+\frac{10}{x-3}$$ and thus, if $M>0$, $$f(x)>M\iff (x-3)(M-6)<10.$$
Here is my problem. If $M>6$, I just set $\delta=\frac{10}{M-6}$ to get $f(x)>M$. But if $M\in (0 ,6]$, which $\delta$ I can take ? 
 A: For $M\leq 6,$ just take, for instance, $\delta=10$. Then $f(x)>7,$ and is in particular greater than $M$. By the way, your if and only if does not hold- $f(x)>M$ if and only if $(x-3)(M-6)<10+(x-3)^2$, which does hold if (but not only if) $(x-3)(M-6)<10$. For instance when $x-3=10, f(x)=17$ is already much bigger than the corresponding $M=7$.
A: Let $M>0$ be given. Set $\delta=\min \{ \frac{1}{M}, 1\}$. So if $|x-3|<\delta$ then
\begin{equation}
\frac{x^2+1}{x-3} \geq \frac{1}{x-3} > M .
\end{equation}
A: Hint:
I give you a proof of the statement, from which you should be able to solve your problem: 
If $x > 3$, then
$$
\frac{x^{2}+1}{x-3} > \frac{x}{x-3} > \frac{3}{x-3};
$$
if $M > 0$, then
$
0 < x-3 < 3/M
$
only if 
$$
\frac{3}{x-3} > M.
$$
We have proved this:
for every $M > 0$, we have $0 < x-3 < 3/M$ only if 
$$
\frac{x^{2}+1}{x-3} > M.
$$
A: For $M<6$, any $\delta>0$ work. Then if you take for all $M>0$, $\delta=\min\left\{\frac{10}{|M-6|},1\right\}$ then $$f(x)>M$$
A: Let $x>3$ and $M>0$.
$$\frac{x^2+1}{x-3}>M\iff x^2+1>M(x-3)$$
You know that $x^2+1>10$ for all $x>3$, then $$x^2+1>M(x-3)\iff M(x-3)<10\iff x-3<\frac{10}{M}$$
Just take $\delta=\frac{10}{M}$, and you'll have $$0<x-3<\delta\implies \frac{x^2+1}{x-3}>M.$$
You don't need other restriction... The only thing is if you would have that your function is defined on $(3,4]$ only. Then you would have the restriction that $3+\delta\leq 4$, and indeed, you can set $\delta=\min\{1,\frac{10}{M}\}$. But even in the case it's not necessary since ou want that $$\forall x\in (3,4], |x-3|<\delta\implies \frac{x^2+1}{x-3}>M$$ and thus, if $\delta=50$ you will naturally have that $x<\delta$ since $x<4$.
