What is the limit of (xcsc2x)/(cos5x) as x approaches 0?

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Homework Statement



lim as x-> 0

(xcsc2x)/(cos5x)


Homework Equations



lim as x-> 0 (sinx)/x = 1


The Attempt at a Solution



(x(1/sin2x)) / (cos5x)

Not sure what I am supposed to do here! Thanks
:)
 
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Firstly, if this is true:

\lim_{x\rightarrow0} \frac{sin(x)}{x} = 1

Then can you figure out what this limit is?

\lim_{x\rightarrow0} \frac{x}{sin(x)}

And then if you use 2x instead of x, this limit should be just as easy:

\lim_{x\rightarrow0} \frac{2x}{sin(2x)}
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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