# What is the limit as x approaches 0 of (2x)/tan(3x)?

Dec 19, 2014

$\frac{2}{3}$

#### Explanation:

The first thing you should always try when calculating limits, is just entering the $x$ value into the function:

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

Seeing that plugging in gives an indeterminate form, we need to use L'Hospital's Rule.

$2 {\lim}_{x \setminus \to 0} \frac{1}{\frac{3 {\csc}^{2} \left(3 x\right)}{\cot} ^ 2 \left(3 x\right)}$

Note that I moved the 2 on top of the limit outside of the limit.

We can move the 3 in the denominator of the 2. $\frac{1}{\csc} ^ 2 \left(x\right)$ is ${\sin}^{2} \left(x\right)$ and $\frac{\frac{1}{1}}{\cot} ^ 2 \left(x\right)$ is simply ${\cot}^{2} \left(x\right)$. But ${\cot}^{2} \left(x\right)$ is ${\cos}^{2} \frac{x}{\sin} ^ 2 \left(x\right)$ so the ${\sin}^{2} \left(x\right)$ gets cancelled leaving:

$\left(\frac{2}{3}\right) {\lim}_{x \setminus \to 0} {\cos}^{2} \left(3 x\right)$.

Cos(0)=0 so we're left with $\frac{2}{3}$ which is our answer.

Feb 18, 2017

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

#### Explanation:

The limit:

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

is of an indeterminate form $\frac{0}{0}$, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then, providing the limits exits then:

${\lim}_{x \rightarrow a} f \frac{x}{g} \left(x\right) = {\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

And so applying L'Hôpital's rule we get:

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

Feb 18, 2017

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

#### Explanation:

Another approach that doesn't rely on using L'Hôpital's rule

We can write the limit as:

${\lim}_{x \rightarrow 0} \frac{2 x}{\tan} \left(3 x\right) = {\lim}_{x \rightarrow 0} \left(2 x\right) \cot \left(3 x\right)$
$\text{ } = 2 {\lim}_{x \rightarrow 0} x \cot \left(3 x\right)$

Let us look at the Taylor Series for $\cot x$, which is as follows:

$\cot x = \frac{1}{x} - \frac{x}{3} - {x}^{3} / 45 - \ldots$

And so we can write a series expansion for the limit:

${\lim}_{x \rightarrow 0} \frac{2 x}{\tan} \left(3 x\right) = 2 {\lim}_{x \rightarrow 0} \left(x\right) \left\{\frac{1}{\left(3 x\right)} - \frac{\left(3 x\right)}{3} - {\left(3 x\right)}^{3} / 45 - \ldots\right\}$
$\text{ } = 2 {\lim}_{x \rightarrow 0} \left\{\frac{1}{3} - {x}^{2} - \frac{27 {x}^{4}}{45} - \ldots\right\}$
$\text{ } = 2 \left(\frac{1}{3}\right)$
$\text{ } = \frac{2}{3}$

Feb 19, 2017

${\lim}_{x \rightarrow 0} \frac{2 x}{\tan} \left(3 x\right) = 2 {\lim}_{x \rightarrow 0} \left(\frac{x}{\sin} \left(3 x\right) \cdot \cos \left(3 x\right)\right)$

$2 {\lim}_{x \rightarrow 0} \frac{3 x}{\sin} \left(3 x\right) \cdot {\lim}_{x \rightarrow 0} \cos \left(3 x\right)$

$= \frac{2}{3} {\lim}_{x \rightarrow 0} \frac{3 x}{\sin} \left(3 x\right) \cdot {\lim}_{x \rightarrow 0} \cos \left(3 x\right)$

$= \frac{2}{3} \left(1\right) \left(1\right) = \frac{2}{3}$

Feb 19, 2017

To determine the limit graphically, se the graph below.

#### Explanation:

graph{2x/tan(3x) [-6.17, 6.316, -2.69, 3.547]}

Zoom in (use wheel) until you can guess that the limit is about $0.667$