# Trig limit question: $\lim_{x\to 0}\frac{1}{2x\csc x}$

How would I solve this limit?

$$\lim_{x\to 0}\frac{1}{2x\csc x}$$

So far this Is what I have done

$\dfrac 1 {2x/ \sin x}= \dfrac {\sin(x)}{2x}$

Would this be then $\dfrac 12 \dfrac{\sin x}x$ ?

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Yes, what you have so far is correct. Do you know what $\displaystyle\lim_{x\to 0}\frac{\sin x}{x}$ is? – Brad Jan 27 '13 at 21:33
it is 1 I see it makes sense – Fernando Martinez Jan 27 '13 at 21:43

Yes, your work is correct thus far. The expression you obtained is perfect. Now it's just a matter of taking its limit as $\,x\to 0$.

The key here is knowing: $\quad\lim_{x \to 0} \dfrac{\sin x}{x} = 1$.

$$\text{Then, }\quad\lim_{x \to 0}\, \frac{1}{2}\cdot\frac{\sin x}{x} \;=\; \frac 12\cdot \lim_{x\to 0}\, \frac {\sin x}{x} \; = \;\frac 12 \cdot 1 \;=\; \frac 12$$

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yes thanks for the help. – Fernando Martinez Jan 27 '13 at 21:53
Your welcome, Fernando! – amWhy Jan 27 '13 at 21:54
Nice! You have a bunch of keys in yor pocket, amWhy. :-) + – S. Snape Jan 28 '13 at 4:25

$$\lim_{x\to 0}\frac{1}{2x\csc x}=\lim_{x\to 0}\frac{1}{2x\frac{1}{\sin x}}=\frac{1}{2}\lim_{x\to 0}\frac{\sin x}{x}=\frac{1}{2}$$

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Nice!, Adi +1 fo rit – S. Snape Jan 30 '13 at 15:07