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I cannot figure out how to solve this trigonometric limit:

$$\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{1}{\tan^2x} \right)$$

I tried to obtain $\frac{x^2}{\tan^2x}$, $\frac{\cos^2x}{\sin^2x}$ and simplify, and so on. The problem is that I always go back to the indeterminate $\infty-\infty$

Has someone a different approach to solve this limit?

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$$ \lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)= \lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^2\sin^2x}= \lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}\frac{x^2}{\sin^2x} $$ Apply l'Hôpital or Taylor expansion (better) to the first fraction.


For the Taylor expansion, it's easier to do it in pieces: $$ \sin x = x-\frac{x^3}{6}+o(x^4),\quad \cos x = 1-\frac{x^2}{2}+o(x^4) $$ so $$ \sin x-x\cos x=x-\frac{x^3}{6}-x+\frac{x^3}{2}+o(x^4)=\frac{x^3}{3}+o(x^4) $$ while $$ \sin x+x\cos x=x-\frac{x^3}{6}+x-\frac{x^3}{2}+o(x^4)=2x+o(x^2) $$ so $$ \lim_{x\to 0}\frac{\sin^2x - x^2\cos^2x}{x^4}= \lim_{x\to 0}\frac{\frac{1}{3}x^3+o(x^4)}{x^3}\frac{2x+o(x^2)}{x}= \lim_{x\to 0}\left(\frac{1}{3}+o(x)\right)(2+o(x))=\frac{2}{3}. $$

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  • $\begingroup$ Thanks for the answer. I have some problems finding the Taylor series, because I need to point it in the $0$, and I found a lot of indeterminate fractions $0/0$ when I try to substitute $x=0$. How could I solve a Taylor series in this case? $\endgroup$ – FdT Nov 28 '13 at 15:17
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Use the series for cotangent $$\cot x = \frac{1}{x}-\frac{1}{3}x-\frac{1}{45}x^3\pm\dots$$ $$\cot^2 x = \frac{1}{x^2}-\frac{2}{3}+\frac{1}{15}x^2\pm\dots $$ $$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)= \lim_{x\to 0}\left(\frac{1}{x^2}-\cot^2 x\right)= \lim_{x\to 0}\left(\frac{2}{3}-\frac{1}{15}x^2 \pm \dots\right) = \frac{2}{3} $$

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  • $\begingroup$ Thanks for your answer. I have a question. How did you apply this series? Is there a way to calculate them? $\endgroup$ – FdT Nov 28 '13 at 15:06
  • $\begingroup$ You can manually use the definition from en.wikipedia.org/wiki/…, but today almost everybody will use a Computer Algebra System (Maple, Maxima etc), or enter e.g. series cot(x)^2 into Wolfram Alpha. $\endgroup$ – gammatester Nov 28 '13 at 15:29
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$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2 x}\right)=\lim_{x\to 0}\frac{\tan x-x}{x^3}\cdot\frac{x+\tan x}{x}\cdot\left(\frac{x}{\tan x}\right)^2= 2\cdot\lim_{x\to 0}\frac{\sec^2 x-1}{3x^2}=$$ $$=\frac{2}{3}\cdot\lim_{x\to 0}\left(\frac{\tan x}{x}\right)^2=\frac{2}{3}.$$

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  • $\begingroup$ Here I used that $\lim_{x\to 0}\cos x=1=\lim_{x\to 0}\frac{\sin x}{x}$ and applied De l'Hopital to $\lim_{x\to 0}\frac{\tan x-x}{x^3}$, using $\frac{d}{dx}\tan x=\sec^2 x=\frac{1}{\cos^2 x}$. No need for the Taylor series of anything. $\endgroup$ – Jack D'Aurizio Nov 28 '13 at 14:10
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$$\lim_{x\to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$$ $$=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^2\tan^2x}\right)$$ $$=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^4}\right)\times \lim_{x\to 0}\left(\frac{x}{\tan x}\right)^2=\lim_{x\to 0}\left(\frac{\tan^2x-x^2}{x^4}\right)$$ Now, using L' Hospital Rule successively for $\frac{0}{0}$ form , $$=\lim_{x\to 0}\left(\frac{2\tan x\sec^2x-2x}{4x^3}\right)=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sec^2x \tan x-x}{x^3}\right)$$ $$=\frac{1}{2}\lim_{x\to 0}\left(\frac{\sec^2x\sec^2x+2\tan^2x\sec^2 x-1}{3x^2}\right)$$ $$=\frac{1}{6}\lim_{x\to 0}\left(\frac{\sec^4x-1+2\tan^2x\sec^2 x}{x^2}\right)$$ $$=\frac{1}{6}\lim_{x\to 0}\left(\frac{(\sec^2x-1)(\sec^2x+1)+2\tan^2x\sec^2 x}{x^2}\right)$$ $$=\frac{1}{6}\lim_{x\to 0}\left(\frac{\tan^2x(\sec^2x+1)+2\tan^2x\sec^2 x}{x^2}\right)$$ $$=\frac{1}{6}\lim_{x\to 0}\left(3\sec^2x+1\right) \times \lim_{x\to 0}\left(\frac{\tan x}{x}\right)^2=\frac{4}{6}\times 1=\frac{2}{3}$$

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