limit problem with M and infinity?

Question

For the limit below, find values of δ that correspond to the M values.
lim tan^2(x) = infinity
(x-->pi/2)

M=1,000
δ=?

Answer 1

*The limit doesn't exist as written. Is this supposed to be the limit from the left? I'll assume so.
*EDIT: Oops, ignore this first part. I missed that it was squared.

What the problem is asking you to do is to find a δ such that f(x) > M = 1000 whenever δ ≤ x < π/2.

Does this help any?

Answer 2

trick in this problem is you have to use some trig identity to get things in terms of (x- pi/2) so that you can solve for (x-pi/2) which gives you the delta in terms of M. To do this you need to use that sine and cosine are 90 degrees off, so tan x is something like cot(x-pi/2) (you check it, maybe negative I don't know).
Finally you take arccot, etc. (delta should be very small)

Answer 3

Well first of all the statement is true because cos(x) x-->pi/2 = 0 and tan^2(x) = sin^2(x) / cos^2(x).
Now do you want to find what value of x (that's supposed to be close enough to 90degrees or pi/2) will skyrocket M to 1000?
Then here is what you do:

tan^2(x) = sin^2(x) / cos^2(x).
Substitute sin^2(x) with the identity1-cos^2(x) and get
(1 - cos^2(x)) / cos^2(x) = 1000
1 / cos^2(x) - cos^2(x) / cos^2(x) = 1000
1 / cos^2(x) = 1001
cos^2(x) = 1/1001
cos(x) = sqrt( 1 / 1001 ) and use your calculator to get ~= 0.0316
Now do inv-cos(0.0316) in radian mode and get 1.539 (in degrees this is 88.something)

I hope I understood your question.