# Question 48bca

Jan 12, 2018

${\lim}_{x \to 0} \frac{\tan 3 x}{2 x} = \frac{3}{2}$

#### Explanation:

Did You mean lim_(x->0)(tan3x)/(2x)?#

If so, then...

Let's rewrite the limit (though not necessary) as

$\frac{1}{2} \cdot {\lim}_{x \to 0} \frac{\tan 3 x}{x}$

If we use direct substitution we will end up with an indeterminate form $\frac{0}{0}$ because:

The ${\lim}_{x \to 0} \tan \left(x\right) = 0$ can easily be proven by looking at its graph:

graph{tanx [-10, 10, -5, 5]}

And plugging in $0$ for $x$ is simply $0$

$\therefore {\lim}_{x \to 0} \frac{\tan \left(3 x\right)}{2 x} = \frac{0}{0}$

Given that, it should become clear that we can apply L'Hospital's Rule which means we take the derivative of both the numerator and denominator separately and then attempt to substitute $0$ again and evaluate the limit.

So,

$\frac{1}{2} \cdot {\lim}_{x \to 0} = \frac{1}{2} \cdot {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \left[\tan \left(3 x\right)\right]}{\frac{d}{\mathrm{dx}} \left[x\right]} = \frac{1}{2} \cdot {\lim}_{x \to 0} \frac{3 {\sec}^{2} \left(3 x\right)}{1}$

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