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.
$\endgroup$ 23 Answers
$\begingroup$$$\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x-\sin x}{x^3}\frac{x+\sin x}{x}\frac{x^2}{\sin^2x}\to\frac{1}{6}\cdot 2\cdot 1$$
Now try the remaining terms.
$\endgroup$ 3 $\begingroup$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 $$\frac1{\sin^2x}-\frac1{x^2}=\frac13+\mathcal O(x^2)$$ $$\lim_{x\to0}\left(\frac1{\sin^2x}-\frac1{x^2}\right)=\frac13$$ $$\lim_{x\to0}\left(\frac1{\sin^2x}+\frac1{\tan^2x}-\frac2{x^2}\right)=\lim_{x\to0}\left(2\left(\frac1{\sin^2x}-\frac1{x^2}\right)-1\right)=2\cdot\frac13-1=-\frac13$$
$\endgroup$ $\begingroup$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 $$ \frac{1}{3}-\frac{2}{3}=-\frac{1}{3} $$
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