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

Expert Answers

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Given to solve,


as `x->0 ` on substituting we get

`arctanx/sinx = 0/0 `

so by using 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)(1/(x^2 +1))/(cosx)`

so now on applying `x->0 ` ie `x=0`

=`(1/(0^2 +1))/(cos0)`


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