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:

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

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

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

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

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

I am aware that $\displaystyle \lim_{x \to 0}\frac{1- cos x}{x} = 0$; 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

Re: Limits with trigonometric functions.

Factor out a $\displaystyle \sin(\theta)$ from both the numerator and denominator of your original ratio, and see where you can go from there.

Re: Limits with trigonometric functions.

I believe you will need to apply L'Hôpital's rule.

$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Re: Limits with trigonometric functions.

$\displaystyle \frac{\tan{x} - \sin{x}}{\sin^3{x}} =$

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

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

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

$\displaystyle \frac{1}{\cos{x}(1+\cos{x})}$

Re: Limits with trigonometric functions.

Quote:

Originally Posted by

**AltF4** I believe you will need to apply L'Hôpital's rule.

$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

why don't you show the OP that method?