Evaluating $ \lim\limits_{x\to 0} \left(\frac{x^4 + 2 x^3 + x^2}{{\tan}^{-1} x}\right)$

Question

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 1

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 :

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

Answer 2

Divide both numerator and denominator by x , and we know $$\lim_{x \to 0 }( (\arctan(x) ) /x ) = 1 $$

Thus the limit becomes $$ \lim_{x \to 0 } x^3 + 2x^2 + x = 0 $$

For the 1st limit, $$ \arctan(x) = x - (1/3) x^3 + (1/5)x^5 +..$$

Thus taking L Hospital here $$ \lim_{x \to 0 }( (\arctan(x) ) /x ) $$ becomes

$$ \lim_{x \to 0 } ( 1 - x^2 ) /1 =1 $$