# `lim_(x->0) cot(2x)sin(6x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply,...

`lim_(x->0) cot(2x)sin(6x)` Find the limit. Use l’Hospital’s Rule where appropriate. If there is a more elementary method, consider using it. If l’Hospital’s Rule doesn’t apply, explain why.

### Textbook Question

Chapter 4, 4.4 - Problem 43 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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lemjay | High School Teacher | (Level 3) Senior Educator

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`lim_(x->0) cot(2x)sin(6x)`

The function cot(2x)sin(6x) is undefined at x=0. So to take its limit, let's apply the L'Hospital's Rule. To do so, express it as a rational function.

`=lim_(x->0) cos(2x)/sin(2x) * sin (6x) = lim_(x->0) (cos(2x)sin(6x))/sin(2x)`

Then, take the derivative of the numerator and denominator.

`=lim_(x->0) ((cos(2x)sin(6x))')/((sin(2x))')=lim_(x->0) (cos(2x)*cos(6x)*6 + (-sin(2x))*2*sin(6x)) / (cos(2x)*2)`

`=lim_(x->0) (6cos(2x)cos(6x) - 2sin(2x)sin(6x))/(2cos(2x))=lim_(x->0) (3cos(2x)cos(6x) - sin(2x)sin(6x))/(cos(2x))`

And, plug-in x=0.

`=(3cos(2*0)cos(6*0)-sin(2*0)sin(6*0))/cos(2*0)=(3*1*1-0)/1=3`

Therefore, `lim_(x->0) cot(2x)sin(6x) = 3` .