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

Answer 1

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#.
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Answer 2

To find the limit of (sin12x)/x as x approaches zero, we can use the limit definition of the derivative. By taking the derivative of sin(12x) with respect to x, we get 12cos(12x). Evaluating this derivative at x=0 gives us 12cos(0) = 12. Therefore, the limit of (sin12x)/x as x approaches zero is 12.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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