Question

What is `lim_(xrarr0+)(ln x) / (csc x)`?

Answer 1

It is `0`.

Explanation:

The initial form is indeterminate `-oo/oo`.

Apply l'Hopital's rule to get

`lim_(xrarr0^+) (1/x)/(-cscxcotx)`

Use trigonometry to rewrite this as

`lim_(xrarr0^+)(sinx/x * sinx/cosx)`

`= 1 * 0/1 = 0`