What is the limit as x approaches 0 of #tan(6x)/sin(2x)#?

Answer 1

The answer is 3:

How did I get there?

The first thing you should always try with limits is just to enter the x value in the function: #lim_{x \to 0}tan(6x)/sin(2x) = tan(6*0)/sin(2*0) = tan(0)/sin(0) = (0/0)#
This is an impossible answer, but whenever we find that we have #(0/0)#, there's a trick we can use. It's called L'Hôpital's Rule. It states formally that: #lim_{x \to a}f(x)/g(x) =lim_{x \to a} (f'(x))/(g'(x))# In this formula, #f'(x)# means the derivative of f(x) and #g'(x)# means the derivative of g(x). If you haven't learned about them yet, you can learn them here.

Let's apply this rule to our problem.

#f(x) = tan(6x)# #f'(x) = d/dx tan(6x) = sec^2(6x)*d/dx(6x)# (chain rule) #= 6*sec^2(6x) = 6/(cos^2(6x))# (since #sec = 1/cos#) So that's our first derivative. Now for our second: #g(x) = sin(2x)# #g'(x) = d/dx sin(2x) = cos(2x)*d/dx(2x)# (chain rule) #= 2*cos(2x)#

Now, all we need to do is combine both of them.

#lim_{x \to 0}tan(6x)/sin(2x) = lim_{x \to 0}((6*sec^2(6x))/(2*cos(2x)))# You can bring the number in front of the limit: #=6/2lim_{x \to 0}((sec^2(6x))/(cos(2x)))# Now we can replace #sec# by #1/cos# #= 3*lim_{x \to 0}(1/(cos^2(6x)*cos(2x)))#
Now let's try entering #x# into this formula: #= 3*(1/(cos^2(6*0)*cos(2*0)))# #= 3*(1/(cos^2(0)*cos(0)))# #= 3*1# (since #cos(0)=1#) #= 3#

And there you have your final answer. Let me know if anything is not clear. I hope this helped.

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Answer 2

#lim_(x->0) (tan 6x) / (sin 2x) = 3#

By de Moivre's theorem:

#cos 3 theta + i sin 3 theta = (cos theta + i sin theta)^3#
#color(white)(cos 3 theta + i sin 3 theta) = cos^3 theta + 3i cos^2 theta sin theta - 3 cos theta sin theta - i sin^3 theta#

Equating imaginary parts:

#sin 3 theta = 3 cos^2 theta sin theta - sin^3 theta#
#color(white)(sin 3 theta) = 3 (1 - sin^2 theta) sin theta - sin^3 theta#
#color(white)(sin 3 theta) = 3 sin theta - 4sin^3 theta#
Putting #theta = 2x# we find:
#lim_(x->0) (tan 6x) / (sin 2x) = lim_(theta->0) (sin 3 theta) / (sin theta cos 3 theta)#
#color(white)(lim_(x->0) (tan 6x) / (sin 2x)) = lim_(theta->0) (3 sin theta - 4 sin^3 theta) / (sin theta cos 3 theta)#
#color(white)(lim_(x->0) (tan 6x) / (sin 2x)) = lim_(theta->0) (3 - 4 sin^2 theta) / (cos 3 theta)#
#color(white)(lim_(x->0) (tan 6x) / (sin 2x)) = (3-0)/1#
#color(white)(lim_(x->0) (tan 6x) / (sin 2x)) = 3#
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Answer 3

The limit as x approaches 0 of tan(6x)/sin(2x) is 3.

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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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