# `lim_(x->0) sin(5x)/tan(9x)` Evaluate the limit, using L’Hôpital’s Rule if necessary.

## Expert Answers

`lim_(x->0) (sin (5x))/(tan(9x))`

To solve, plug-in x = 0.

`lim_(x->0) (sin(5*0))/(tan(9*0)) = 0/0`

Since the result is indeterminate, to find the limit of the function as x approaches zero, apply L'Hopital's Rule. To do so, take the derivative of the numerator and denominator.

`lim_(x->0) (sin(5x))/(tan(9x)) = lim_(x->0) ((sin(5x))')/((tan(9x))') = lim_(x->0)...

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`lim_(x->0) (sin (5x))/(tan(9x))`

To solve, plug-in x = 0.

`lim_(x->0) (sin(5*0))/(tan(9*0)) = 0/0`

Since the result is indeterminate, to find the limit of the function as x approaches zero, apply L'Hopital's Rule. To do so, take the derivative of the numerator and denominator.

`lim_(x->0) (sin(5x))/(tan(9x)) = lim_(x->0) ((sin(5x))')/((tan(9x))') = lim_(x->0) (5cos(5x))/(9sec^2(5x))`

And, plug-in x =0.

`= (5cos(5*0))/(9sec^2(5*0) )= (5cos(0))/(9sec^2(0)) =(5*1)/(9*1)=5/9`

Therefore,  `lim_(x->0) (sin (5x))/(tan(9x))=5/9` .

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