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`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 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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