What is the limit of # (1-cos(x))/sin(x^2) # as x approaches 0?

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Answer 1 · 2016-07-08T23:58:54

#lim_(x->0) (1-cos(x))/(sin(x^2))=0/0# (an indeterminate form), thus we can apply

L'Hospital's rule, since substituting #x=0# yields #0/0#.

In this case, applying L'Hospital's rule we get

#lim_(x->0) (sin(x))/(2x * cos(x^2)) -> 0/0#, which also yields an indeterminate form,

thus we apply it again:

#lim_(x->0) (cos(x))/(2cos(x^2)-4x^2 * sin(x^2))#

#=lim_(x->0) (cos(x))/(2(cos(x^2)-2x^2 * sin(x^2))#

#=cos(0)/(2(cos(0)-2(0)(sin(0))#

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

L'Hospital's rule states that if function(s) #f# and #g# are differentiable on an open interval #I# except at #c# contained in #I#, then if

#lim_(x->c) f(x) = lim_(x->c) g(x) = 0# or #± ∞#,

#g'(x) ≠ 0# for all # x in I# with #x ≠ c#, and

#lim_(x->c) (f'(x)) / (g'(x))# exists, then

#lim_(x->c) (f(x)) / (g(x)) = lim_(x->c) (f'(x)) / (g'(x))#

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