Limit questions

Question

Evaluate the limit..

Lim _x-->0    1-cos 5x / 1-cos 7x

 

Answer 1

Method I. Using L'Hopital's rule,

lim{x-->0} (1-cox5x)/(1-cos7x)

= lim{x-->0} (5sin5x)/(7sin7x)

= 25/49

 

Method II. Using double angle formula, 1- cos(2A) = 2sin^2(A)

lim{x-->0} (1-cox5x)/(1-cos7x)
= lim{x-->0} 2sin^2(5x/2)/[2sin^2(7x/2)]

= 25/49

 

Answer 2

Let us first simplify the equation.

(1-cox5x)/(1-cos7x) .... multiply numerator and denominator by (1+cos5x)(1+cos7x)

=[(1-cos5x)(1+cos5x)(1+cos7x)]/[(1-cos7x)(1+cos7x)(1+cos5x)]......remember that (a+b)(a-b)=a²-b²

=[(1-cos²5x)(1+cos7x)]/[(1-cos²7x)(1+cos5x)]  ....1-cos²x = sin²x

=[sin²5x(1+cos7x)]/[sin²7x(1+cos5x)]....rearrange and multiply and divide by 25x².49x²

=[25x².sin²5x/25x²][49x²/(49x²sin²7x)][(1+cos7x)/(1+cos5x).....rearrange

=(sin5x/5x)².(7x/sin7x)2.(25x²/49x²).(1+cos7x)/(1+cos5x)

now simplify and take limit. Remember that limit _x->0 (sinx/x) = 1

=(25/49) . lim _x->0 (sin5x/5x)² . lim _x->0 (7x/sin7x)² . lim _x->0 (1+cos7x)/(1+cos5x)

=(25/49).1.1.(1+cos0)/(1+cos0) = (25/49).(2/2)

=25/49