# Trigonometry limit problem?

How would I find the following limit: $$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}$$

I am not sure how to do this one help will be appreciated.

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L'Hopital's rule is the most usual method for this sort of thing. –  Michael Hardy Jan 26 '13 at 22:15
I do not know that rule... –  Fernando Martinez Jan 26 '13 at 22:31
I am starting to understand thanks for the answers. –  Fernando Martinez Jan 26 '13 at 22:33

Multiply top and bottom by $1+\cos(3x)\;$ (which is called the conjugate of $1 - \cos(3x))$.

$$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2} \cdot \frac{1 + \cos(3x)}{1 + \cos(3x)}$$ $$= \lim_{x \to 0} \frac{1 - \cos^2(3x)}{2x^2(1 + \cos(3x)}$$ $$= \lim_{x\to 0}\; \frac{\sin^2(3x)}{2x^2(1 + \cos(3x))}$$ $$=\lim_{x\to 0} \frac{\sin^2(3x)}{3^2\cdot x^2}\cdot \frac{3^2}{2[1+\cos(3x)]}$$ $$=\lim_{x\to 0} \left(\frac{\sin(3x)}{3x}\right)^2\frac{9}{2(1+\cos(3x))}$$ $$=\lim_{x\to 0} \left(\frac{\sin(3x)}{3x}\right)^2\cdot \frac{9}{2}\cdot \frac{1}{(1+\cos(3x))}=1\cdot\frac{9}{2\cdot 2} = \frac94$$

I'm assuming you know the value of $\lim_{x\to 0}\dfrac{\sin ax}{ax} = 1\,$ where $a$ is a nonzero constant.

If you know L'hospital, you can apply that twice (differentiate each of the numerator and denominator twice), and evaluate the resulting limit. If you haven't learned it yet, you will likely learn it soon, and it can greatly simplify problems like this!

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No I have not learned it yet,I have just started my calc class. –  Fernando Martinez Jan 26 '13 at 22:36
No problem...I filled in some more details (not using L'hospital), just a trig identity ($1 - \cos^2(3x) = \sin^2(3x)$) and the fact that $\lim_{x\to 0} \dfrac{\sin (ax)}{ax} = 1$ ($a$ a constant not equal to $0$) –  amWhy Jan 26 '13 at 22:42
One question I have is when you have (9/2) (sin3x/3x)^2(1+cos3x) would not have (9/2)(1)(2) and this would be 18/2=9 ? –  Fernando Martinez Jan 26 '13 at 22:48
No, there's only one factor of $2$ in the original denominator, and that is moved over to the right fraction (since it's all being multiplied.) I added a line: Note that the factor $(1 + \cos(3x)) \to (1 + 1) = 2$ as $x \to 0$, since $\cos(3\cdot 0) = 1$ –  amWhy Jan 26 '13 at 22:54
Yes I see now I got confused for a second thanks. –  Fernando Martinez Jan 26 '13 at 22:57

Hint: Multiply top and bottom by $1+\cos(3x)$. Then the limit should follow from what you know about $\lim_{t\to 0}\frac{\sin t}{t}$.

Or else more mechanically use L'Hospital's Rule twice.

Remark: The mathematically most natural approach is none of the above. By the usual power series expansion for $\cos t$, we have $$1-\cos(3x)=\frac{(3x)^2}{2!}-\frac{(3x)^4}{4!}+\cdots.$$ Divide the right side by $2x^2$, and let $x\to 0$.

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so would I have I have sin^2(3x)/(2x^(2)(1+cos(3x) –  Fernando Martinez Jan 26 '13 at 22:25
As a first step you get $\frac{\sin^2(3x)}{2x^2(1+\cos(3x))}$. Rewrite this as $\frac{9}{2(1+\cos(3x))}\left(\frac{\sin(3x)}{3x}\right)^2$. –  André Nicolas Jan 26 '13 at 22:34

$$1- \cos(3x) = 2 \sin^2(3x/2)$$ Hence, $$\dfrac{1-\cos(3x)}{2x^2} = \dfrac{\sin^2(3x/2)}{x^2} = \left(\dfrac{\sin(3x/2)}x\right)^2$$ Now recall that $\displaystyle \lim_{t \to 0}\dfrac{\sin(at)}t = a$ and if $\lim_{t \to b} f(t) = \tilde{f} \in \mathbb{R}$, then $\lim_{t \to b} f^2(t) = \tilde{f}^2$

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$$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}=\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}\frac{1+\cos(3x)}{1+\cos(3x)}=\lim_{x\to 0} \frac{1-\cos^2(3x)}{2x^2}\frac{1}{1+\cos(3x)}=$$ $$=\lim_{x\to 0} \frac{\sin^2(3x)}{2{(3x)}^2}\frac{9}{1+\cos(3x)}=\lim_{x\to 0} \left(\frac{\sin(3x)}{3x}\right)^2\frac{9}{2(1+\cos(3x))}=\frac{9}{4}$$ Using L'Hopital rule $$\lim_{x\to 0} \frac{1-\cos(3x)}{2x^2}=\lim_{x\to 0} \frac{3\sin(3x)}{4x}=\lim_{x\to 0} \frac{9\cos(3x)}{4}$$

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How is (1+cos3x) equal to two is because cosx=1 as x approach zero? –  Fernando Martinez Jan 26 '13 at 22:35
Yes, that's it. –  André Nicolas Jan 26 '13 at 22:36
+1 Nice approach –  Babak S. Jan 27 '13 at 15:35