# `lim_(x->0) x/arctan(2x)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

Given to solve,

`lim_(x->0) x/arctan(2x)`

as `x->0` then the `x/arctan(2x) =0/0` form

so upon applying the L 'Hopital rule we get the solution as follows,

as for the general equation it is as follows

`lim_(x->a) f(x)/g(x) is = 0/0` or `(+-oo)/(+-oo)` then by using the L'Hopital Rule we get  the...

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Given to solve,

`lim_(x->0) x/arctan(2x)`

as `x->0` then the `x/arctan(2x) =0/0` form

so upon applying the L 'Hopital rule we get the solution as follows,

as for the general equation it is as follows

`lim_(x->a) f(x)/g(x) is = 0/0` or `(+-oo)/(+-oo)` then by using the L'Hopital Rule we get  the solution with the  below form.

`lim_(x->a) (f'(x))/(g'(x))`

so , now evaluating

`lim_(x->0) x/arctan(2x)`

=`lim_(x->0) (x')/((arctan(2x))')`

= `lim_(x->0) 1/((arctan(2x))')`

First let us compute the

`(arctan(2x))' `

let `u= 2x`

=>

so,

`(arctan(2x))'= d/dx (arctan(2x)) `

=`d/(du) (arctan(u)) d/dx (u)`             [as `d/dx f(u) = d/(du) f(u) * d/dx (u)]`

= `(1/(u^2+1) d/dx (2x)`

=`(1/(u^2+1) )(2)`

=`(2/(u^2+1))` but u= `2x` ,so

=`2/((2x)^2+1)`

= `(2/(4(x)^2+1)) `

now coming back to the limits , we have

`lim_(x->0) 1/(arctan(2x))'`

= `lim_(x->0) 1/(2/(4x^2+1))`

as x->0 , we get

=`1/((2/(4(0)^2+1)) )`

=` 1/2`

so , we can state that ,

`lim_(x->0) x/arctan(2x) = 1/2`

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