# `lim_(x->0) sin(3x)/sin(5x)` Evaluate the limit, using L’Hôpital’s Rule if necessary. `lim_(x->0) (sin(3x))/(sin(5x))`

To solve, plug-in x=0.

`lim_(x->0) (sin(3x))/(sin(5x)) = (sin(3*0))/(sin(5*0)) = 0/0`

Since the result is indeterminate, to solve for the limit of the function as x approaches zero, apply the L'Hopital's Rule. So, take the derivative of the numerator and denominator.

`lim_(x->0) (sin(3x))/(sin(5x))= lim_(x->0) ((sin(3x))')/((sin(5x))')= lim_(x->0) (3cos(3x))/(5cos(5x))`

And, plug-in...

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`lim_(x->0) (sin(3x))/(sin(5x))`

To solve, plug-in x=0.

`lim_(x->0) (sin(3x))/(sin(5x)) = (sin(3*0))/(sin(5*0)) = 0/0`

Since the result is indeterminate, to solve for the limit of the function as x approaches zero, apply the L'Hopital's Rule. So, take the derivative of the numerator and denominator.

`lim_(x->0) (sin(3x))/(sin(5x))= lim_(x->0) ((sin(3x))')/((sin(5x))')= lim_(x->0) (3cos(3x))/(5cos(5x))`

And, plug-in x = 0.

`= (3cos(3*0))/(5sin(5*0)) = (3*1)/(5*1)=3/5`

Therefore,  `lim_(x->0) (sin(3x))/(sin(5x))=3/5` .

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