Showing that a function diverges for large x Consider the function $$f(x) = \frac{x^{3}}{x^{2}+1}$$ Show that as $x\to\infty$ we have that $f(x)\to\infty$. ie. I want to prove that $$\lim_{x\to\infty} \frac{x^3}{x^2+1} = \infty.$$
So given $k>0$ I want to find $M>0$ such that $$x>M\implies |f(x)|>k.$$
First note that 
\begin{align}
|f(x)| &= \left|\frac{x^3}{x^2+1}\right|\\
&\ge\left|\frac{x^3}{2x^2}\right|\\
&= \frac{x}{2}
\end{align}
So to ensure $|f(x)|>k$, we can let $x/2>k$, or $x>2k$, thus whenever $M=2k$ we are sure that for all $x>M$ 
\begin{align}
|f(x)|&=\left|\frac{x^3}{x^2+1}\right|\\
&\ge\left|\frac{x^3}{2x^2}\right|\\
&= \frac{x}{2} \\
&> k.& \text{since } x>M>0\implies \frac{x}{2}>\frac M2=k>0.
\end{align}
Is this a sufficient and correct proof?

Write $f(x) = x-{x \over x^2+1}$, and since $|{x \over x^2+1}| \le {1 \over 2}$, you have
$f(x) \ge x-{1 \over 2}$. Then for all $k>0$, if you let $M=k+1$, then you have the desired result.
