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How to calculate: $$\lim \limits_{x \to 0}\frac{1}{x^2} - \frac{1}{\sin^2x} $$

any suggestions what I can do here?

Aaron Maroja
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Firas Abd El Gani
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2 Answers2

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$$-\frac{1}{\sin(x)^2} + \frac{1}{x^2} = \frac{-x^2+\sin(x)^2}{x^2 \sin(x)^2}.$$

You can use L'Hopital's rule now, but it will take multiple applications since both the numerator and denominator are very small (of order $x^4$, it turns out) near $0$.

A nicer approach is to use the Maclaurin series for $\sin$ to see that

$$\sin(x)^2=(x-x^3/6+O(x^5))(x-x^3/6+O(x^5))=x^2-x^4/3+O(x^6).$$

Consequently

$$-x^2+\sin(x)^2=-x^4/3+O(x^6)$$

and

$$x^2\sin(x)^2=x^4+O(x^6).$$

Now take the ratio and cancel a factor $x^4$ from top and bottom to compute the limit.

Ian
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    You can reduce the work by first multiplying by $\frac{\sin^2x}{x^2}$. Then by factoring the denominator and using l'hoptal, or something else on each of the two factors separately – Rene Schipperus Feb 13 '15 at 23:57
  • True; $\frac{\sin(x)^2}{x^2} \left ( \frac{1}{x^2} - \frac{1}{\sin(x)^2} \right ) = \frac{\sin(x)^2}{x^4} - \frac{1}{x^2} = \frac{\sin(x)^2-x^2}{x^4}$. So that saves the work of showing that $x^2 \sin(x)^2$ is equivalent to $x^4$ on this scale. – Ian Feb 14 '15 at 00:53
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    further $\frac{\sin x+x}{x}\to 2$ is clear, so one is left with $\frac{\sin x -x}{x^3}$ and one aplication of l'hopital gives another well known limit. – Rene Schipperus Feb 14 '15 at 01:01
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[If allowed to use Taylor series] Hint: $\sin^2 x \sim x^2$

Alex
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