What is the limit as x approaches 0 of $\frac{x}{arctan (4 x)}$ $x/arctan(4x)$ ?

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Answer 1 · 2018-07-16T16:49:50

$lim_(xto0) x/(arc tan4x)=1/4$

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 · 2018-07-16T16:55:40

$1/4$

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

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

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

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