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

2 Answers
Dec 19, 2014

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.

Oct 12, 2016

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

Explanation:

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#