`y=csc(2x)` Find the limit, if possible

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Chapter 3, 3.9 - Problem 14 - Calculus of a Single Variable (10th Edition, Ron Larson).
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If you evaluate the limit of function from the left, as `x -> 0^-` yields:

`lim_(x->0^-) csc(2x) = lim_(x->0^-) 1/(sin(2x))`

Replacing `2sinx*cosx` for `sin(2x)` yields:

`lim_(x->0^-) 1/(sin(2x)) = lim_(x->0^-) 1/(2sinx*cosx)`

`lim_(x->0^-) 1/(2sinx*cosx) = (1/2)lim_(x->0^-) 1/(sin x)*lim_(x->0^-) 1/(cos x)`

Replacing `0^-` for x yields:

`(1/2)lim_(x->0^-) 1/(sin x)*lim_(x->0^-) 1/(cos x) = (1/2)*(1/(sin 0^-))*1/(cos 0^-)`

`(1/2)lim_(x->0^-) 1/(sin x)*lim_(x->0^-) 1/(cos x) = (1/2)*(-oo)*1 = -oo`

If you evaluate the limit of function from the right, as `x -> 0^+` yields:

`lim_(x->0^+) csc(2x) = lim_(x->0^+) 1/(sin(2x))`

`lim_(x->0^+) 1/(sin(2x)) = (1/2)*(1/(sin 0^+))*1/(cos 0^+) = (1/2)*(+oo)*1 = +oo`

Evaluating the limit of `csc(2x)` as x approaches to `0^-` yields` -oo` , while evaluating the limit of `csc(2x)` as x approaches to `0^+` yields +oo, hence the two sided limit does not exist.

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