# How do you find the limit of (tan 3x)^x as x approaches 0?

Jul 13, 2016

1

#### Explanation:

${\lim}_{x \to 0} {\left(\tan 3 x\right)}^{x}$

$= {\lim}_{x \to 0} {e}^{\ln \left({\left(\tan 3 x\right)}^{x}\right)}$

$= \exp \left({\lim}_{x \to 0} x \ln \left(\left(\tan 3 x\right)\right)\right)$ as e^x is continuous function we
can lift it outside the limit

$= \exp \left({\lim}_{x \to 0} \frac{\ln \left(\left(\tan 3 x\right)\right)}{\frac{1}{x}}\right)$ which is indeterminate so we can use LHospital's rule

$= \exp \left({\lim}_{x \to 0} \frac{\frac{3 {\sec}^{2} 3 x}{\left(\tan 3 x\right)}}{\frac{1}{x} ^ 2}\right)$

$= \exp \left({\lim}_{x \to 0} \frac{1}{\cos} ^ 2 \left(3 x\right) \cdot \frac{3 {x}^{2}}{\tan 3 x}\right)$ and we can lift the cos term out as continuous and apply L'Hopital again

$= \exp \left(\frac{1}{\cos} ^ 2 \left(3 x\right) \cdot {\lim}_{x \to 0} \frac{6 x}{3 {\sec}^{2} 3 x}\right)$

$= \exp \left(\frac{1}{\cos} ^ 2 \left(3 x\right) \cdot {\lim}_{x \to 0} \left(6 x\right) \cdot \left(3 {\cos}^{2} 3 x\right)\right)$

$= \exp \left(1 \cdot 0 \cdot 3\right) = 1$