Question

How do you find the limit of `sintheta/(theta+tantheta)` as `theta->0`?

Answer 1

`= 1/2`

Explanation:

`lim_(theta to 0) sintheta/(theta+tantheta)`

...is `0/0` indeterminate, so L'Hôpital's Rule appies...

`= lim_(theta to 0) (cos theta)/(1+sec^2 theta)`

`= lim_(theta to 0) (cos theta)/(1+1/(cos^2 theta))`

...all of which functions are continuous at the limit

`= lim_(theta to 0) (1)/(1+1/(1)) = 1/2`