Find the points of discontinuity: $f(x) = (x^4+x^3+2x^2)/\tan^{-1}(x)$ if $x\ne0$ and $f(0)=10$ Here is the question:

Find the points of discontinuity:
  $$f(x) = 
\begin{cases}
\frac{x^4+x^3+2x^2}{\tan^{-1}x} & \text{if} \ x\ne0 \\ \\ 
10 & \text{if} \ x=0 \\
\end{cases}.$$

$\frac{x^4+x^3+2x^2}{\tan^{-1}x}$ being the division of a polynomial function and a inverse trigonometric function (both continuous) is continuous at each $x\ne0$ and $10$ being a constant function is also continuous.
At $x=0,$
$$\lim_{x\to 0}\frac{x^4+x^3+2x^2}{\tan^{-1}x}=?$$
How do I simplify this?

Applying L'Hospital:
$$\lim_{x\to 0}\frac{x^4+x^3+2x^2}{\tan^{-1}x}=\lim_{x\to 0}\frac{4x^3+3x^2+4x}{\frac{1}{1+x^2}}=0\ne 10$$
Therefore, $f$ is not continuous at $0$.
