Limit of trigonometric equation

I'm working through some trig questions and don't understand this one:

$\displaystyle \lim_{ x \to 0} \frac{1 - \cos 2x}{3x^{2}}$

My first thought was since $\displaystyle \lim_{x \to 0} \cos x = 1$, the answer would be 0, but alas, it turned out to be $\displaystyle \frac{2}{3}$

Could anyone shed some light on this?

Re: Limit of trigonometric equation

Quote:

Originally Posted by

**Lepzed** I'm working through some trig questions and don't understand this one:

$\displaystyle \lim_{ x \to 0} \frac{1 - \cos 2x}{3x^{2}}$

My first thought was since $\displaystyle \lim_{x \to 0} \cos x = 1$, the answer would be 0, but alas, it turned out to be $\displaystyle \frac{2}{3}$

Note that $\displaystyle \cos (2x)=1-2\sin^2(x)$ thus

$\displaystyle \frac{1 - \cos (2x)}{3x^{2}}=\frac{2\sin^2(x)}{3x^{2}}=\frac{2}{3 }\left(\frac{\sin(x)}{x}\right)^2$

Re: Limit of trigonometric equation

Ah, so I should've rewritten the cos 2x expression, before applying the limit.

Thanks! Really helpful!

Re: Limit of trigonometric equation

I worked through another exercise and I'm wondering if my approach is correct. It might be a bit verbose, but I'd like to be clear on my train of thought:

$\displaystyle \lim_{x \to 0} \frac{x^2}{\sin x \tan 3x} = \lim_{x \to 0} \frac{x^2}{\sin x \frac{\sin 3x}{\cos 3x}} = \lim_{x \to 0} \frac{x^2}{\frac{ \sin x \cos x \sin 3x}{\cos 3x}} = \lim_{x \to 0} \frac{x^2 \cos 3x}{\sin x \cos x \sin 3x} = \lim_{x \to 0} \frac{x \cos 3x}{ \sin 3x \cos x } = \lim_{x \to 0} \frac{x}{ \sin 3x} = \frac{1}{3}$

Am I doing anything illegal here? I can't really identify with the meaning of this, so it's a mechanical process for me, I can't visualize this at all.

Re: Limit of trigonometric equation

Quote:

Originally Posted by

**Lepzed** I worked through another exercise and I'm wondering if my approach is correct. It might be a bit verbose, but I'd like to be clear on my train of thought:

$\displaystyle \lim_{x \to 0} \frac{x^2}{\sin x \tan 3x} = \lim_{x \to 0} \frac{x^2}{\sin x \frac{\sin 3x}{\cos 3x}} = \lim_{x \to 0} \frac{x^2}{\frac{ \sin x \cos x \sin 3x}{\cos 3x}} = \lim_{x \to 0} \frac{x^2 \cos 3x}{\sin x \cos x \sin 3x} = \lim_{x \to 0} \frac{x \cos 3x}{ \sin 3x \cos x } = \lim_{x \to 0} \frac{x}{ \sin 3x} = \frac{1}{3}$

It would easier to follow writing it as:

$\displaystyle \left( {\frac{x}{{\sin (x)}}} \right)\left( {\frac{x}{{\sin (3x)}}} \right)\left( {\frac{{\cos (3x)}}{{\cos (x)}}} \right)$