# Limits with trigonometric functions.

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• Sep 16th 2012, 02:17 PM
amillionwinters
Limits with trigonometric functions.
I am having trouble with this limit. I don't need an answer, but I would appreciate if someone could give me a hint or show me what I am missing.

The following is what I can make sense of it so far:

$\lim_{x \to 0}\frac{tan\theta - sin\theta}{sin^3\theta}$

$\rightarrow \lim_{x \to 0}\frac{\frac{sin\theta}{cos\theta} - sin\theta}{sin^3\theta}$

$\rightarrow \lim_{x \to 0}\left(\frac{sin\theta}{cos\theta}\right)\left( \frac{1}{sin^3\theta}\right) - \frac{sin\theta}{sin^3\theta}$

$\rightarrow \lim_{x \to 0}\frac{1}{cos\theta cos^2\theta} - \frac{1}{sin^2\theta}$

$\rightarrow \lim_{x \to 0}\frac{1- cos\theta}{cos\theta sin^2\theta}$

I am aware that $\lim_{x \to 0}\frac{1- cos x}{x} = 0$; however, I am a little lost on how to get there, or if that is even the direction I should be taking. Any hints or explanations would be greatly appreciated. Thanks!

Take care.
/alan
• Sep 16th 2012, 02:29 PM
SworD
Re: Limits with trigonometric functions.
Factor out a $\sin(\theta)$ from both the numerator and denominator of your original ratio, and see where you can go from there.
• Sep 16th 2012, 02:34 PM
AltF4
Re: Limits with trigonometric functions.
I believe you will need to apply L'Hôpital's rule.
$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$
• Sep 16th 2012, 02:47 PM
skeeter
Re: Limits with trigonometric functions.
$\frac{\tan{x} - \sin{x}}{\sin^3{x}} =$

$\frac{\sin{x} - \cos{x}\sin{x}}{\cos{x}\sin^3{x}} =$

$\frac{1 - \cos{x}}{\cos{x}\sin^2{x}} =$

$\frac{1 - \cos{x}}{\cos{x}(1-\cos^2{x})} =$

$\frac{1}{\cos{x}(1+\cos{x})}$
• Sep 16th 2012, 03:16 PM
skeeter
Re: Limits with trigonometric functions.
Quote:

Originally Posted by AltF4
I believe you will need to apply L'Hôpital's rule.
$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

why don't you show the OP that method?