Limit of trig functions

Question

What is the limit

lim x ->0 sin x^2 /(1-cos x)

Answer 1

Another approach is to use Pythagorean trig identity and difference of squares:

lim_x->0 (sin²(x))/(1 - cos(x))

= lim_x->0 (1 - cos²(x))/(1 - cos(x))

=lim_x->0 [(1 - cos(x))(1 + cos(x))]/(1 - cos(x))

= lim_x->0 1 + cos(x) = 1 + 1 = 2.

Answer 2

Use the identity sin^2(x)= 1- cos^2(x)

1-cos^2(x)/ ( 1-cosx) Now factor the numerator as the difference of squares

(1+cos(x)) ( 1-cos(x) ) / (1-cosx) this simplifies to ( 1+cosx)

The limit of this expression is 1 + cos(0)

Cos(0)= 1

so you have 1+1 =2

Answer 3

Both the numerator and denominator equal 0 when x = 0. You can either use l'Hopital's rule twice or calculate the Maclaurin series for the numerator and denominator.

Also, if you substitute .001 for x in the two expressions, you can know what your result should be.

sin(x^2)' = 2x cos(x^2) Note that this equals 0 when x = 0

(2x cos(x^2))' = 2cos(x^2)-4x^2 sin(x^2). This equals 2 when x = 0

(1 - cos(x))' = sin(x). This equals 0 when x = 0.

(sin(x))' = cos(x). This equals 1 when x = 0

sin(x^2)'' /(1 - cos(x))'' = (2 cos(x^2) - 4x^2 sin(x^2))/cos(x), which has a limit of 2/1 = 2.