How do you find the limit of #(x(1-cosx))/tan^3x# as #x->0#?

Question

4957 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-02T00:04:20

#(x(1-cosx))/tan^3x# as #xrarr0#

Substitution yields the indeterminate form #0/0#

#(x(1-cosx))/tan^3x = (x(1-cosx)cos^3x)/(sin^3x)#

# = x/sinx * (1-cosx)/sin^2x * cos^3x#

The limit that is still a problem is the middle limit.

#(1-cosx)/sin^2x = (1-cosx)/sin^2x * (1+cosx)/(1+cosx)#

# = (1-cos^2x)/(sin^2x(1+cosx))#

# = sin^2x/(sin^2x(1+cosx))#

# = 1/(1+cosx)#

#lim_(xrarr0)(x(1-cosx))/tan^3x = lim_(xrarr0)(x/sinx * 1/(1+cosx) * cos^3x)#

# = (1)(1/(1+1))(1^3) = 1/2#