Solving (cot x) ^ sin2x at the Limit of x→0

[teng125]

anybody knows how to solve this using L'hospital rule pls
(cot x) ^ sin2x with limi X to zero

 

[TD]

You can write:

f\left( x \right)^{g\left( x \right)} = e^{\ln \left( {f\left( x \right)^{g\left( x \right)} } \right)} = e^{g\left( x \right)\ln \left( {f\left( x \right)} \right)}

So:

\mathop {\lim }\limits_{x \to 0} f\left( x \right)^{g\left( x \right)} = e^{\mathop {\lim }\limits_{x \to 0} \left( {g\left( x \right)\ln \left( {f\left( x \right)} \right)} \right)}

Does this help you any further? Remember that for L'Hopital, you're trying to get 0/0 or ∞/∞.