Proving that $\lim \limits_{x \to \infty}\frac{x^3}{x^2 +10}=\infty$ using $\epsilon-\delta$ definition $\lim \limits_{x \to \infty}\frac{x^3}{x^2 +10}=\infty$
Therefore $\forall M>0$ we need to find a $\delta=\delta(M)$ such that $f(x)>M$ or $x<\delta(M)$
so $\frac{x^2}{x^2+10}>M$
$x^2>M(x^2+10)$
$x^2-Mx^2>10M$
$x^2(1-M)>10M$
$x>\sqrt{\frac{10}{1-M}}$
But this doesn't help me because x is greater than the function $\delta(M)$

The limit
$$
\lim_{x\to \infty} \frac{x^2}{x^2+10} = 1
$$
and not equal to $\infty$. You can see this from
$$
\lim_{x\to \infty} \frac{x^2 / x^2}{(x^2+10)/x^2} = \lim_{x\to \infty} \frac{1}{1 + 10/x^2} = 1.
$$

If you are considering the limit
$$
\lim_{x\to \infty} \frac{x^3}{x^2+10} = \lim_{x\to \infty} \frac{x}{1 + 1/x^2}
$$
then let $M$ be given. You want to find a $\delta$ such that if $x > \delta$, then $\frac{x}{1 + 1/x^2} > M$. Now let $\delta_1>0$ be  be such that $\frac{1}{x^2} < 1$ for $\delta > 0$. You probably know this exists because $\lim_{x\to \infty} \frac{1}{x^2} = 1$. So you have $1+\frac{1}{x^2} < 2$ for $x > \delta_1$.
Let $\delta_2 > 2M$. Then let $\delta = \max\{\delta_1, \delta_2\}$. This will work.
