# trig limits

• Nov 8th 2013, 02:20 PM
Jonroberts74
trig limits
Evaluate $\displaystyle \lim_{x\rightarrow 0} \frac{1-\cos4x}{x^2}$

so by the double angle

$\displaystyle \cos 4x = \cos^2 2x - \sin^2 2x$

this I understand but it shows it go further to $\displaystyle 1-2\sin^2 2x$

This is where it starts to lose me.
• Nov 8th 2013, 02:36 PM
topsquark
Re: trig limits
Quote:

Originally Posted by Jonroberts74
Evaluate $\displaystyle \lim_{x\rightarrow 0} \frac{1-\cos4x}{x^2}$

so by the double angle

$\displaystyle \cos 4x = \cos^2 2x - \sin^2 2x$

this I understand but it shows it go further to $\displaystyle 1-2\sin^2 2x$

This is where it starts to lose me.

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

You can find these by taking the cos^2 - sin^2 expression by applying $\displaystyle sin^2(x) + cos^2(x) = 1$

-Dan
• Nov 8th 2013, 09:56 PM
Jonroberts74
Re: trig limits
okay so

if $\displaystyle \cos^2 x = 1 - \sin^2 x$ true, then is $\displaystyle \cos^2 2x = 1 - \sin^2 2x$ true? That is the way my text book makes it seem

it skips over this step

$\displaystyle 1 - \sin ^2 2x - \sin^2 2x$

then inputing that back in

I get

$\displaystyle \lim_{x\rightarrow 0} \frac{1- \cos 4x}{x^2} = \lim_{x \rightarrow 0} \frac{1 - (1 - \sin 2 sin^2\,\, 2x)}{x^2} \Rightarrow \lim_{x\rightarrow0} \frac{2 \sin^2 2x}{x^2}$

The work the book shows then gets a bit confusing from here (I am working on reviewing key trig components so please bear with me, the book is not always very clear.)
• Nov 9th 2013, 05:54 PM
Prove It
Re: trig limits
Quote:

Originally Posted by Jonroberts74
Evaluate $\displaystyle \lim_{x\rightarrow 0} \frac{1-\cos4x}{x^2}$

so by the double angle

$\displaystyle \cos 4x = \cos^2 2x - \sin^2 2x$

this I understand but it shows it go further to $\displaystyle 1-2\sin^2 2x$

This is where it starts to lose me.

Since this is of the indeterminate form \displaystyle \displaystyle \begin{align*} \frac{0}{0} \end{align*}, L'Hospital's Rule can be applied.

\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{1 - \cos{(4x)}}{x^2} &= \lim_{x \to 0}\frac{\frac{d}{dx}\left[ 1 - \cos{(4x)} \right] }{\frac{d}{dx} \left( x^2 \right) } \\ &= \lim_{x \to 0} \frac{ 4\sin{(4x)} }{ 2x} \\ &= \lim_{x \to 0} \frac{2\sin{(4x)}}{x} \end{align*}

This is again of the form \displaystyle \displaystyle \begin{align*} \frac{0}{0} \end{align*}, so we can use L'Hospital's Rule again...

\displaystyle \displaystyle \begin{align*} \lim_{ x \to 0 } \frac{2\sin{(4x)}}{x} &= \lim_{x \to 0} \frac{\frac{d}{dx} \left[ 2\sin{(4x)} \right] }{\frac{d}{dx}\left( x \right) } \\ &= \lim_{x \to 0} \frac{8\cos{(4x)}}{1} \\ &= \lim_{x \to 0} 8\cos{(4x)} \\ &= 8\cos{(4 \cdot 0 )} \\ &= 8\cdot 1 \\ &= 8 \end{align*}
• Nov 9th 2013, 06:21 PM
Jonroberts74
Re: trig limits
I don't know that rule yet but I know it comes up shortly for me so I will return to this then, thank you!