Question

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

Answer 1

`lim_(xto0) x/(arc tan4x)=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_(xto0) x/(arc tan4x)`

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

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

`=>L=lim_(yto0)(1/4tany)/y`

`=>L=1/4lim_(yto0)(siny/cosy)/y`

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

`L=1/4(1)(1/cos0)`

`L=1/4`

Answer 2

`1/4`

Explanation:

Since `lim_{xto 0}(4x)/(arctan(4x))=1` we get

`lim_{xto 0}x/arctan(4x)=1/4`
By L'Hospital we get

`lim_(xto 0) (x)/arctan(4x)=lim_(x to 0)4/((1/(1+16x^2))*4)=lim_(x to 0)1+16x^2=1`