1- Can anyone tell me how to solve the limits required for question 3.a Section D Problem Set 1

6 tan(3x)(sec(3x)^2) = 6 sin (x)/ (cos(x)^3)

2 - How do the 3's before the x's cut?

Thanks in advance.

Question

1- Can anyone tell me how to solve the limits required for question 3.a Section D Problem Set 1

6 tan(3x)(sec(3x)^2) = 6 sin (x)/ (cos(x)^3)

2 - How do the 3's before the x's cut?

Thanks in advance.

calculusfunctions · Answer

Is your question:

6tan(3x)sec²(3x) = (6sinx)/cos³x

or

6tan(3x)sec(3x)² = (6sinx)/cos(x)³ ?

Because there is fundamental difference between those two equations. When you don't express yourself clearly, it makes it difficult for people to help you.

anonymous · Answer

What is the difference?

anonymous · Answer

He is asking if you are cubing/squaring the result you get after taking the cos of the angle or are you taking the cos of the cube of the angle. (I am pretty sure hes asking:
6tan(3x)sec²(3x) = (6sinx)/cos³x

anonymous · Answer

Yes, that's the equation van1234. The cube is outside the cos function, as I wrote. Any ideas how the 3's inside the tan and sec functions cut? Btw any insight on the first question? Thanks.

anonymous · Answer

Is this the question you are referring to in your first question:

$$(x-2)\div(x ^{2}-4)$$

If so, then here's my solution.
You know that the denominator is a difference of squares, so you can factor that to be:

(x+2) (x-2)

Then you see that the x-2 's cancel, (keep in mind though that you will have a hole at x = 2, as x=2 still does not work in the original function). You are left with:

$$1 \div (x +2)$$
Which means that there is an asymptote at x = -2.

anonymous · Answer

Thanks van1234, any idea about the second question?