Find the value of #lim_(xto0)(1-cos(ax))/(1-cos(bx))#?

1 Answer
Mar 8, 2018

#lim_(xto0)(1-cos(ax))/(1-cos(bx))=>lim_(xto0)(asin(ax))/(bsin(bx))=a^2/b^2#

Explanation:

By using L'Hopital's rule, we know that If #(f(a))/(g(a))=0#, then #lim_(xtoa)(f(x))/(g(x))=(f'(a))/(g'(a))#

We have #lim_(xto0)(1-cos(ax))/(1-cos(bx))#

#(d/dx[1-cos(ax)])/(d/dx[1-cos(bx)])=(asin(ax))/(bsin(bx))#

However, #(asin(0a))/(bsin(0b))=0/0="undefined"# so wee must do another limit.

#lim_(xto0)(asin(ax))/(bsin(bx))=>(d/dx[asin(ax)])/(d/dx[bsin(bx)])=(a^2cos(ax))/(b^2cos(bx))#

#(a^2cos(0b))/(b^2cos(0b))=a^2/b^2#

#lim_(xto0)(1-cos(ax))/(1-cos(bx))=>lim_(xto0)(asin(ax))/(bsin(bx))=a^2/b^2#