Evaluate $\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)$

Question

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

Answer 1

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 $x \to 0$)

Answer 2

L'Hospital:

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

Answer 3

I prefer to apply elementary trigonometry before getting to the Taylor expansion, in order to avoid dealing with the expansion of $\sin^2x.$ $${1\over\sin^2x}-{1\over x^2}= {2x^2-2\sin^2x\over 2x^4}{x^2\over \sin^2x}$$ Next $$2x^2-2\sin^2x=2x^2-1+\cos(2x)\\ ={(2x)^4\over 4!}+o(x^6)={2\over 3}x^4+o(x^6) $$ Therefore $${1\over\sin^2x}-{1\over x^2}={1\over 3}+o(x^2)$$