# Question 56924

Jan 16, 2017

$\text{ The Limit=} \frac{3}{8}$.

#### Explanation:

Let us use the following Standard Forms of Limit :

 (1) lim_(xrarr0)sinx/x=1, &, (2) lim_(xrarr0)tanx/x=1#

Hence, Reqd. Lim. $= {\lim}_{x \rightarrow 0} \frac{\frac{\tan 3 x}{3 x} \left(3 x\right)}{\frac{\sin 8 x}{\left(8 x\right)} \left(8 x\right)}$

$= {\lim}_{x \rightarrow 0} \frac{\frac{\tan 3 x}{3 x}}{\frac{\sin 8 x}{8 x}} \left(\frac{3}{8}\right)$

$= \left(\frac{1}{1}\right) \left(\frac{3}{8}\right)$

$\therefore \text{ The Limit=} \frac{3}{8}$.

Jan 16, 2017

When ${\lim}_{x \to 0} \tan \frac{3 x}{\sin} \left(8 x\right)$ is evaluated at 0, we obtain $\frac{0}{0}$; this means that L'Hôpital's rule should be used.

#### Explanation:

To use L'Hôpital's rule:

1. Compute the derivative of the numerator: $\frac{d \left(\tan \left(3 x\right)\right)}{\mathrm{dx}} = 3 {\sec}^{2} \left(3 x\right)$

2. Compute the derivative of the denominator: $\frac{d \left(\sin \left(8 x\right)\right)}{\mathrm{dx}} = 8 \cos \left(8 x\right)$

3. Write the limit with the new numerator and the new denominator: ${\lim}_{x \to 0} \frac{3 {\sec}^{2} \left(3 x\right)}{8 \cos \left(8 x\right)}$ This limit can be found by evaluation at 0:

${\lim}_{x \to 0} \frac{3 {\sec}^{2} \left(3 x\right)}{8 \cos \left(8 x\right)} = \frac{3}{8}$

L'Hôpital's rule states that; as this limit goes, so goes the original limit:

${\lim}_{x \to 0} \tan \frac{3 x}{\sin} \left(8 x\right) = \frac{3}{8}$