How do you evaluate the limit #(1-cosx)/x# as x approaches #0#?

2 Answers
Douglas K.
Sep 26, 2016

Because the expression evaluated at the limit is #0/0#, we can use L'Hospital's Rule.

Explanation:

The rule states that, when given the limit of a fraction, #(lim)/(xrarrc) f(x)/g(x)# where #f(c)/g(c)# is, #0/0#, #oo/oo# or #(-oo)/-oo#, then #(lim)/(xrarrc)(f'(x))/(g'(x))# goes to the same limit.

To implement the rule, we take the derivative of the numerator:
#d(1 - cos(x))/dx = sin(x)#

the derivative of the of the denominator:

#dx/dx = 1#

Assemble the new fraction with the same limit:

#(lim)/(xrarr0) sin(x)/1 = 0#

Therefore, #(lim)/(xrarr0) (1 - cos(x))/x = 0#

Jim H
Sep 26, 2016

Usually, this is done after showing that #lim_(xrarr0)sinx/x = 1#. See below.

Explanation:

#(1-cosx)/x = ((1-cosx))/x * ((1+cosx))/((1+cosx))#

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

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

# = sinx/x * sinx * 1/(1+cosx)#

Taking the limit as #xrarr0#, we get

#(1)(0)(1/2) = 0#

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