Limit of a trigonometric thing

Jan Hill
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Homework Statement


Find the limit as x approaches zero of xcsc4x = the numerator over cos17x = the denominator


Homework Equations





The Attempt at a Solution



I don't know how to start except maybe multiplying both numerator and denominator by 4
 
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This?
\lim_{x \to 0} \frac{x~csc(4x)}{cos(17x)}

Start with some trig identities, such as csc(x) = 1/sin(x). There are some trig limits that will be helpful, such as lim sin(x)/x = 0, as x --> 0.
 
so does this become lim as x approaches 0 of x times 1/sin(4x)
and this is lim as x approaches 0 of x/sin(4x)

and what trig limit should I use to deal with the denominator?


and please accept my many thanks for your quick reply of earlier!
 
Jan Hill said:
so does this become lim as x approaches 0 of x times 1/sin(4x)
and this is lim as x approaches 0 of x/sin(4x)
That's part of it.

It would be helpful if you wrote actual mathematics equations and expressions, rather than verbose English descriptions.
Jan Hill said:
and what trig limit should I use to deal with the denominator?


and please accept my many thanks for your quick reply of earlier!
 
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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