Limit of trigonometric equation

**Lepzed** · July 9th 2012, 11:17 AM

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

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

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

Could anyone shed some light on this?

**Plato** · July 9th 2012, 11:49 AM

Originally Posted by Lepzed

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

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

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

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

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

**Lepzed** · July 9th 2012, 12:13 PM

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

Thanks! Really helpful!

**Lepzed** · July 10th 2012, 07:23 AM

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:

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

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.

**Plato** · July 10th 2012, 07:46 AM

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:

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

It would easier to follow writing it as:
$\left( {\frac{x}{{\sin (x)}}} \right)\left( {\frac{x}{{\sin (3x)}}} \right)\left( {\frac{{\cos (3x)}}{{\cos (x)}}} \right)$

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