Question

How do you solve `\lim _ { x \rightarrow 0} ( \frac { \tan 3x } { \sin 2x } )`?

Answer 1

Because evaluation at the limit gives the indeterminate form `0/0`, one should use L'Hôpital's rule

Explanation:

`lim_(xrarr0) tan(3x)/sin(2x)`

Using L'Hôpital's rule :

`lim_(xrarr0) ((d(tan(3x)))/(dx))/((d(sin(2x)))/(dx))`

`lim_(xrarr0) (3sec^2(3x))/(2cos(2x))`

`lim_(xrarr0) 3/2sec^3(3x)= 3/2`

Therefore, so goes the original limit:

`lim_(xrarr0) tan(3x)/sin(2x) = 3/2`