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I tried l'Hospital but that will require a lot (and I mean A LOT!!!) of differentiating

Is there a shortcut? $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$$

Thanks in advance

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It should only require two or three derivatives. – Gaffney Jan 6 '14 at 17:00
You could rewrite $\sin^2x$ to its third order Taylor expansion. – Ragnar Jan 6 '14 at 17:01
@Ragnar unfortunately, we skipped Taylor so I am not allowed to use it on an exam. Thanks, though. – Ahmed Ali Jan 6 '14 at 17:05
you can use the limit $\lim_{x \rightarrow 0} \frac{ \sin x}{x}=1$and making some algebraic manipulations – twin prime Jan 6 '14 at 17:15
up vote 2 down vote accepted

Of course there is!

$$\sin x \sim x - \frac{x^3}{6}$$

$$\sin^2 x \sim x^2 - \frac{x^4}{3}$$

So $$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$$ $$= \lim_{x \to 0} \frac{x^2 - \sin^2 x}{x^2 \cdot \sin^2 x} = \lim_{x \to 0} \frac{\frac{x^4}{3}}{x^4 - \frac{x^6}{3}} = \frac{1}{3}$$

(cause also $x^4 \pm x^6 \sim x^4$ if $x \to 0$)

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$$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right) =\mathop {\lim }\limits_{x \to 0} \frac{x^2-\sin^2 x}{x^2\sin^2x}=\mathop {\lim }\limits_{x \to 0}\frac{x+\sin x}{x}\cdot\frac{x^2}{\sin^2x}\cdot\frac{x-\sin x}{x^3}=2\cdot1\cdot\mathop {\lim }\limits_{x \to 0}\frac{(x-\sin x)'}{(x^3)'}=2\mathop {\lim }\limits_{x \to 0}\frac{1-\cos x}{3x^2}=\frac{2}{3}\cdot\frac{(1-\cos x)'}{(x^2)'} =\frac{2}{3}\cdot\frac{\sin x}{2x}=\frac{1}{3}$$

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