# What is the limit as x approaches 0 of x/arctan(4x)?

Jul 16, 2018

${\lim}_{x \to 0} \frac{x}{a r c \tan 4 x} = \frac{1}{4}$

#### Explanation:

We know that ,

color(red)((1)lim_(theta to0)(sintheta)/theta=1

color(red)((2)lim_(theta to0) costheta=1

We take ,

$L = {\lim}_{x \to 0} \frac{x}{a r c \tan 4 x}$

Let , color(blue)(arctan4x=y=>4x=tany=>x=1/4tany

=>color(blue)(xto0 =>y=arctan(0)=>yto0

$\implies L = {\lim}_{y \to 0} \frac{\frac{1}{4} \tan y}{y}$

$\implies L = \frac{1}{4} {\lim}_{y \to 0} \frac{\sin \frac{y}{\cos} y}{y}$

=>L=1/4lim_(yto0)siny/y*1/(lim_(yto0)cosy)to color(red)(Apply(1)and(2)

$L = \frac{1}{4} \left(1\right) \left(\frac{1}{\cos} 0\right)$

$L = \frac{1}{4}$

Jul 16, 2018

$\frac{1}{4}$

#### Explanation:

Since ${\lim}_{x \to 0} \frac{4 x}{\arctan \left(4 x\right)} = 1$ we get

${\lim}_{x \to 0} \frac{x}{\arctan} \left(4 x\right) = \frac{1}{4}$
By L'Hospital we get

${\lim}_{x \to 0} \frac{x}{\arctan} \left(4 x\right) = {\lim}_{x \to 0} \frac{4}{\left(\frac{1}{1 + 16 {x}^{2}}\right) \cdot 4} = {\lim}_{x \to 0} 1 + 16 {x}^{2} = 1$