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I have to compute

$$ \lim_{x\to0^+}\frac{\pi/2- \arctan(1/x^2)-\sin(x^3)-1+\cos(x)}{x\tan(x)+e^{x^3}-1} $$

I separated the numerator so I got that

$$\dfrac{-1+\cos(x)}{x\tan(x)+e^{x^3}-1} \longrightarrow -\dfrac{1}2;$$

I know that the limit is $\dfrac{1}2$ and I know, by checking on Wolfram Alpha, that

$$\dfrac{-\sin(x^3)}{x\tan(x)+e^{x^3}-1}\longrightarrow 0$$ so

$$\dfrac{π/2- \arctan(1/x^2)}{x\tan(x)+e^{x^3}-1} \longrightarrow 1.$$

I tried using L'Hopital but it gets even more complicated. How can I solve it?

Lorenzo B.
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  • use power series instead of L'Hopital's rule (you have many examples here - just open some of "Related" links), e.g. http://math.stackexchange.com/questions/548832/how-find-this-limit-lim-x-to-0-dfrac-sin-tanx-tan-sinxx?rq=1 – larry01 Feb 14 '17 at 10:21
  • I have not studied power series yet – Lorenzo B. Feb 14 '17 at 10:22
  • Start using $\frac{\pi }{2}-\tan ^{-1}\left(\frac{1}{x^2}\right)=\tan ^{-1}\left({x^2}\right)$ – Claude Leibovici Feb 14 '17 at 10:28
  • i think you might be able to use l'hopital twice – Cato Feb 14 '17 at 10:28
  • @ClaudeLeibovici, Thank you, I did not know I could do that.

    I think it works now: $x\tan(x)$ tends to $x^2$ and $e^{x^3}-1$ tends to 0.

    The numerator tends to $x^2$ so it becomes $x^2/x^2$ which is $1$. Am I right?

     – Lorenzo B. Feb 14 '17 at 10:30
  • I got a 1/2 using l'hopital twice, and confirmed by putting small values into a computer – Cato Feb 14 '17 at 10:56

2 Answers2

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Remember that, for $t>0$, $$ \arctan t+\arctan\frac{1}{t}=\frac{\pi}{2} $$ so we can write $$ \frac{\pi}{2}-\arctan\frac{1}{x^2}=\arctan(x^2) $$ This simplifies things a bit.

Now prove that $$ \lim_{x\to0^+}\frac{x\tan x+e^{x^3}-1}{x^2}=1 $$ which follows from $$ \lim_{x\to0^+}\frac{x\tan x}{x^2}=1 $$ and from $$ \lim_{x\to0^+}\frac{e^{x^3}-1}{x^2}= \lim_{x\to0^+}x\frac{e^{x^3}-1}{x^3}=0\cdot 1=0 $$ Thus the limit you have to compute is the easier $$ \lim_{x\to0^+}\frac{\arctan(x^2)-\sin(x^3)-1+\cos x}{x^2}= \lim_{x\to0^+} \left(\frac{\arctan(x^2)}{x^2}-\frac{\sin(x^3)}{x^2}-\frac{1-\cos x}{x^2}\right) $$

egreg
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differentiate numerator

$d/dx(\pi/2 - \arctan(\frac{1}{x^2}) - \sin(\frac{1}{x^3}) - 1 + \cos(x)) $

$= -[(-2) / x^3] / [1 - (\frac{1}{x^2})^2] - 3x^2\cos(x^2) - \sin(x)$

$= 2x / [x^4 + 1] - 3x^2\cos(x^2) - \sin(x)$

you can differentiate that again, but i think that the only non zero term at x = 0 will be $2x / [x^4 +1] - \cos(x)$ - giving the value 1 for the numerator after two differentiations

for the denominator, the first differentiation is

$\tan(x) + x\sec^2(x) + 3x^2 \exp(x^3)$

a second differentiation of that, and setting x =0 is going to give the non-zero terms

$\sec^2(x) + \sec^2(x) + x(\sec(x) \tan(x)) + [zero terms]$

= $\sec^2(x) + \sec^2(x) = 2$

so I make it $\frac{1}{2}$

also I got it as 1/2 as an estimate on a computer, via numerical method

Cato
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