limits - Evaluating $limlimits_{xto 0} left(frac{x^4 + 2 x^3 + x^2}{{tan}^{-1} x}right)$

In a question from a class test, we are given this function:
$$f(x) = \begin{cases} \frac{x^4 + 2 x^3 + x^2}{{\tan}^{-1} x}, & \text{if $ x \neq 0$} \\[2ex] 0, & \text{if $x = 0$} \end{cases}$$
We are asked to find whether $f(x)$ is continuous at $x=0$

.

Now, we can get the solution by Taylor expansion or L'Hopital's rule quite easily.
But, L'Hopital's rule and Taylor expansions aren't a part of my course syllabi this year so I don't think they need to be applied here.

But I can't figure out how to evaluate this:
$$\lim_{x \to 0} \left(\frac{x^4 + 2 x^3 + x^2}{{\tan}^{-1} x}\right)$$ without these methods.

I think the first step should be factorizing the numerator to get
$$f(x) = \frac {x^2(x+1)^2}{{\tan}^{-1}x}$$

Now I don't know how to proceed further.
Is there some identity that can be used here?

Answer

With the derivative :

$\displaystyle \lim_{x\to 0}\frac{\arctan(x)- \arctan(0)}{x-0}=f'(0)=\dfrac{1}{1+(0)^2}=1\iff \displaystyle \lim_{x\to 0}\frac{\arctan(x)}{x}=1$

Thus :

$\displaystyle \lim_{x \to 0} \dfrac{x^4 + 2 x^3 + x^2}{\arctan x}=\lim_{x \to 0} \dfrac{x^3 + 2 x^2 + x}{\frac{\arctan(x)}{x}}=0$