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

1 Answer
Jun 26, 2016

#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]}