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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?

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    $\begingroup$ 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 $\endgroup$ – Ahmed S. Attaalla Sep 3 '17 at 5:05
  • $\begingroup$ @Ahmed S. Attaalla, oh okay. Thanks for explaining! $\endgroup$ – Mr Reality Sep 3 '17 at 5:10
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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 :

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

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

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

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  • $\begingroup$ 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?) $\endgroup$ – Mr Reality Sep 3 '17 at 4:56
  • $\begingroup$ I edited my answer. Hope you get it. $\endgroup$ – Sitanshu Sep 3 '17 at 5:00
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    $\begingroup$ 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). $\endgroup$ – Ahmed S. Attaalla Sep 3 '17 at 5:02
  • $\begingroup$ @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? $\endgroup$ – Mr Reality Sep 3 '17 at 5:03
  • $\begingroup$ @Ahmed S. Attaalla Thanks. Will do that. Sorry for the trouble. $\endgroup$ – Sitanshu Sep 3 '17 at 5:05

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