Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
1  
It should only require two or three derivatives. –  Gaffney Jan 6 at 17:00
2  
You could rewrite $\sin^2x$ to its third order Taylor expansion. –  Ragnar Jan 6 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 at 17:05
2  
you can use the limit $\lim_{x \rightarrow 0} \frac{ \sin x}{x}=1$and making some algebraic manipulations –  twin prime Jan 6 at 17:15
add comment

2 Answers 2

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 $x \to 0$)

share|improve this answer
add comment

L'Hospital:

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.