Solve using L'hospital rule.
Lim as x goes to 0 Cot2xSin6x

Question

Solve using L'hospital rule.
Lim as x goes to 0 Cot2xSin6x

psymon · Answer

You have to put the function into an indeterminant form first. This can be done simply rewriting cot2x as cos2x/sin2x. Then you can use l'hopitals just fine. Just make sure to know that l'hopitals does the derivative of the numerator separate from the derivative of the denominator, not together. So we would have (cos2xsin6x) on top to differentiate first. 

-2sin2xsin6x + 6cos2xcos6x all divided by the derivative of sin2x, leaving us with:

$$\frac{ -2\sin2xsin6x+6\cos2xcos6x }{2\cos2x }$$Now that we have cosine in the denominator, we don't have to worry about an undefined answer. So plugging in x = 0 we have:

$$\frac{ (0)(0) +6(1)(1) }{ 2(1) }= 3$$

anonymous · Answer

Thank you so much!!! I do not know why that was so hard for me.

psymon · Answer

Usually people forget it has to be indeterminant form or they try doing quotient rule xD As long as you understand it :3

anonymous · Answer

You are great. I wish I could keep you in reserve :)