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Find the limit $$\lim_{x\to 0} \cos \bigg(\pi x^2 \csc (\frac {x} {2}) \cot (6x) \bigg)$$

I dont even know where to get started... Some hints and solutions would be appreciated! Thanks in advance!

P.S typed this on an iphone, sorry for any mistakes will edit soon.

EDIT

Here are my current workings.

$$\lim_{x\to 0} \cos \bigg(\pi x^2 \csc (\frac {x} {2}) \cot (6x) \bigg)=\lim_{x\to 0} \cos \bigg(\pi x^2 (\frac {\cos (6x)} {\sin (\frac {x} {2}) \sin 6x} \bigg)$$

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5 Answers 5

up vote 1 down vote accepted

Note that $$\pi x^2 \csc (\frac {x} {2}) \cot (6x)=\frac{\pi}{3}\cdot\frac{\frac{x}{2}}{\sin(\frac{x}{2})}\cdot\frac{6x}{\sin(6x)}\cos(6x).$$ Since $\lim_{x\to 0}\frac{\sin x}{x}=1$ and $\lim_{x\to 0}\cos(x)=1$, we have $$\lim_{x\to 0}x^2 \csc (\frac {x} {2}) \cot (6x)=\frac{1}{3}.$$ Since $\cos$ is a continuous function, we have $$ \lim_{x\to 0}\cos \Big(\pi x^2 \csc (\frac {x} {2}) \cot (6x) \Big) =\cos \Big(\lim_{x\to 0}\pi x^2 \csc (\frac {x} {2}) \cot (6x) \Big)=\cos\frac{\pi}{3}=\frac{1}{2}.$$

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Quite a hint${}$ –  Gerry Myerson Sep 27 '12 at 5:08

HINT:

$$\begin{align*} \lim_{x\to 0}~\cos\left(\pi x^2 \csc \left(\frac {x} {2}\right) \cot (6x)\right)&=\lim_{x\to 0}~\cos\left(\frac{\pi x^2}{\sin(x/2)\tan 6x}\right)\\ &=\lim_{x\to 0}~\cos\frac{\pi}3\left(\frac{x/2}{\sin(x/2)}\cdot\frac{6x}{\tan 6x} \right)\\ &=\cos\frac{\pi}3\left(\lim_{x\to 0}\frac{x/2}{\sin(x/2)}\right)\left(\lim_{x\to 0}\frac{6x}{\tan 6x}\right)\;, \end{align*}$$

since the cosine is a continuous function. The limits in the last line are ones that you should know.

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$$\pi x^2\csc\left(\frac{x}{2}\right)\cot(6x)=\pi x^2\cdot\frac{2}{x}\cdot\frac{1}{6x}+\ldots=\frac{\pi}{3}+O(x)$$ so the limit is $\cos(\pi/3)=\frac{1}{2}$.

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OP may have to work harder to fill in the steps here than to solve the problem originally. –  Gerry Myerson Sep 27 '12 at 4:56
    
I dont really understand the csc and cot to a bunch of x and ... What formula is that? Thanks anyway! –  Yellow Skies Sep 27 '12 at 4:59
    
Ok im working it out. Ill post my steps soon –  Yellow Skies Sep 27 '12 at 5:00
    
To leading order, $\sin x\approx x$ and $\cos x\approx 1$. –  Jonathan Sep 27 '12 at 5:17

Hints: express all the trig functions in terms of the sine and the cosine, and then manipulate things so you can use what you might know about $\lim_{x\to0}((\sin x)/x)$.

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$$\begin{align*} \lim_{x\to 0} \cos (\pi x^2 \csc (\frac {x} {2}) \cot (6x)) &=\cos(\lim_{x\to 0}\pi x^2 \csc (\frac {x} {2}) \cot (6x))\\ &=\cos(\lim_{x\to 0}\frac{\pi}{3}.\frac{\frac{x}{2}}{\sin\frac{x}{2}}.\frac{6x}{\sin6x}.\cos6x)\\ &=\cos(\frac{\pi}{3}.\lim_{x\to 0}\frac{\frac{x}{2}}{\sin\frac{x}{2}}.\lim_{x\to 0}\frac{6x}{\sin6x}.\lim_{x\to 0}\cos6x)\\ &=\cos\frac{\pi}{3}=\frac{1}{2} \end{align*}$$

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That's a hint?$ $ –  Gerry Myerson Sep 27 '12 at 5:08

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