Question

How do you evaluate `(csc(x) - cot(x))` as x approaches 0?

Answer 1

`Lt_(x->0)(cscx-cotx)=0`

Explanation:

`cscx-cotx=1/sinx-cosx/sinx=(1-cosx)/sinx`

Now series expansion of `sinx` and `cosx` is

`sinx=x-x^3/(3!)+x^5/(5!)-x^7/(7!)+..................`
`cosx=1-x^2/(2!)+x^4/(4!)-x^6/(6!)+..................`

Hence `1-cosx=x^2/(2!)-x^4/(4!)+x^6/(6!)-..................`

and `Lt_(x->0)(cscx-cotx)`

= `Lt_(x->0)(x^2/(2!)-x^4/(4!)+x^6/(6!)-..................)/(x-x^3/(3!)+x^5/(5!)-x^7/(7!)+..................)`

= `Lt_(x->0)(x/(2!)-x^3/(4!)+x^5/(6!)-..................)/(1-x^2/(3!)+x^4/(5!)-x^7/(7!)+..................)=0/1=0`

graph{(cscx-cotx) [-10, 10, -5, 5]}