# Difficult limit problem involving sine and tangent

I encountered the following problem:

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

I have tried to separate it into two limits (one with sine and the other with tangent) and applied L'Hôpital's rule, but even third derivative doesn't work.

I also tried to simplify the expression a bit:

$$\frac 1{\sin^2 x} + \frac 1{\tan^2 x} = \frac{1+\cos^2 x}{\sin^2 x} = \frac{ 1}{1-\cos x} + \frac 1{1+\cos x} -1$$

But I cannot make it work either. I would like answers with series expansion. Thanks in advance.

• reduction of fractions to a common denominator? – W. mu Oct 20 '16 at 4:13
• note $1+\cos^2 x= 2-\sin^2 x$. You essentially need to compare $1/x^2$ with $2/ \sin^2 x$ – Thomas Oct 20 '16 at 4:15

$$\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x-\sin x}{x^3}\frac{x+\sin x}{x}\frac{x^2}{\sin^2x}\to\frac{1}{6}\cdot 2\cdot 1$$

Now try the remaining terms.

• Ah, thats clever. Thanks for the answer. May I know how did you come up with such a decomposition? – lEm Oct 20 '16 at 4:29
• Its not that clever, you add the fractions, and get a difference of squares. The second idea is to convert $\sin$ into $x$, as in the solution. – Rene Schipperus Oct 20 '16 at 4:33
• I see, I had tried to break it into $\frac{x+\sin x}{x\sin x}$ and $\frac {x-\sin x}{x\sin x}$ but it turns out one of the limit doesn't exist. For some reasons I haven't thought of separating the $x$ from $\sin x$ – lEm Oct 20 '16 at 4:44

An alternative approach which does use the series expansion asked for in the question proceeds as follows. $$\frac1{\sin^2x}+\frac1{\tan^2x}-\frac2{x^2}$$ $$=\frac{1+\cos^2x}{\sin^2x}-\frac2{x^2}$$ $$=\frac{2-\sin^2x}{\sin^2x}-\frac2{x^2}$$ $$=\frac2{\sin^2x}-1-\frac2{x^2}$$ $$=2\left(\frac1{\sin^2x}-\frac1{x^2}\right)-1$$ The Laurent series of $\frac1{\sin^2x}$ around zero is $$\frac1{x^2}+\frac13+\mathcal O(x^2)$$ (see e.g. here for the derivation). Therefore $$\frac1{\sin^2x}-\frac1{x^2}=\frac13+\mathcal O(x^2)$$ $$\lim_{x\to0}\left(\frac1{\sin^2x}-\frac1{x^2}\right)=\frac13$$ $$\lim_{x\to0}\left(\frac1{\sin^2x}+\frac1{\tan^2x}-\frac2{x^2}\right)=\lim_{x\to0}\left(2\left(\frac1{\sin^2x}-\frac1{x^2}\right)-1\right)=2\cdot\frac13-1=-\frac13$$

Since $\sin x$ and $\tan x$ are “almost equal” to $x$, separating into a sum of two limits seems like an interesting approach. The first limit is $$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)= \lim_{x\to0}\frac{x^2-\sin^2x}{x^2\sin^2x}$$ Finding a suitable Taylor expansion of the numerator is easy: $$x^2-\sin^2x= (x-\sin x)(x+\sin x)= \Bigl(\frac{x^3}{6}+o(x^3)\Bigr)\bigl(2x+o(x)\bigr)=\frac{1}{3}x^4+o(x^4)$$

For the second limit you don't even need to remember the Taylor expansion of the tangent (but it's easier if you do): $$\lim_{x\to0}\left(\frac{1}{\tan^2x}-\frac{1}{x^2}\right)= \lim_{x\to0}\frac{x^2\cos^2x-\sin^2x}{x^2\sin^2x}$$ Now $$(x\cos x-\sin x)(x\cos x+\sin x)= \Bigl(x-x\frac{x^2}{2}-x+\frac{x^3}{6}+o(x^3)\Bigr) \bigl(2x+o(x)\bigr)=-\frac{2}{3}x^4+o(x^4)$$ Hence your limit is $$\frac{1}{3}-\frac{2}{3}=-\frac{1}{3}$$