How do you find the limit of #(sin12x)/x# as x approaches zero?

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Answer 1 · 2015-08-09T22:08:46

Use L'Hôpital's rule with #f(x) = sin(12x)# and #g(x) = x#.

#f'(x) = 12 cos 12x# and #g'(x) = 1#, so

#lim_(x->0) f(x)/g(x) = lim_(x->0) (f'(x))/(g'(x)) = lim_(x->0) (12 cos 12x)/1 = 12#

L'Hôpital's rule says that if:

(1) #f(x)# and #g(x)# are differentiable on an open interval #I# containing #c#, except possibly at #c# and

(2) #lim_(x->c) f(x) = lim_(x->c) g(x) = 0# and

(3) #lim_(x->c) (f'(x))/(g'(x))# exists and #g'(x) != 0# for all #x in I# (except possibly at #c#)

then

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

With #f(x) = sin(12x)# and #g(x) = x# these 3 conditions hold for #c = 0#.