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

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?

• With $x=\tan u$ we get $\lim_{x \to 0} \frac{\arctan x}{x}=\lim_{u \to 0} \frac{u}{\tan u}=\lim_{u \to 0} \frac{1}{\frac{\sin u}{u}} \cos u=1$. @Mr Reality – Ahmed S. Attaalla Sep 3 '17 at 5:05
• @Ahmed S. Attaalla, oh okay. Thanks for explaining! – Mr Reality Sep 3 '17 at 5:10

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 :

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

• Thanks for your answer. But this answer uses the same method for solving the problem as the previous answer. – Mr Reality Sep 3 '17 at 7:32
• But I don't use L'Hopital rule!!! – Stu Sep 3 '17 at 7:42
• I use the definition of a derivative :$$\lim_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}=f'(x_0)$$ – Stu Sep 3 '17 at 7:48
• Okay, I think I didn't read it carefully enough. – Mr Reality Sep 3 '17 at 8:28
• +1 ( I would surely have accepted this answer if you had written it earlier) ;) – Mr Reality Sep 3 '17 at 9:31

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

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

• Ok, I did not know about the identity for arctan(x) that you used. Do you know of any source when I can read up on a simple proof of this identity? (I see it can be proved by applying L'Hopital's rule: can that be taken as a valid proof?) – Mr Reality Sep 3 '17 at 4:56
• I edited my answer. Hope you get it. – Sitanshu Sep 3 '17 at 5:00
• I suggest you learn how to typeset if you haven't already @Sitanshu. To type $\lim_{x \to a} f(x)$ write \lim_{x \to a} f(x). – Ahmed S. Attaalla Sep 3 '17 at 5:02
• @Ahmed S. Attaalla, Then I just get $\frac{u}{tan (u)}$. In a sense that's just the numerator and denominator interchanged, right? What do I need to do now? – Mr Reality Sep 3 '17 at 5:03
• @Ahmed S. Attaalla Thanks. Will do that. Sorry for the trouble. – Sitanshu Sep 3 '17 at 5:05