Limit problem?

Question

this is a limit problem. The problem is:
lim sec(x)-tan(x) / pi/2 - (x)
x-->pi/2

Please give answer with explaination.

Answer 1

Dutch Prof is right.

Answer 2

I assume the expression in the limit is

[sec x - tan x] / [pi/2 - x]

Write the secant and tangent as fractions involving cosine: sec x = 1/cos x, tan x = sin x/cos x. Then we have

[1 - sin x] / [cos x * (pi/2 - x)]

We can replace x = pi/2 - u and take the limit for u -> 0 of

[1 - cos u] / [u * sin u]

We can expand cos u = 1 - u^2/2 + ... and sin u = u - u^3/6 + ..., so the limit becomes

[u^2/2] / [u^2 - u^4/6 + ...] = 1/2

Answer 3

Well I'm at work so I'm not really prepared to think but it doesn't look like this is possible. Tan(pi/2) DNE and we can't use L'hopital's rule because derivative of sec(x) doesn't help the problem not to mention I don't think you could subtract the two "infinities" in the numerator to make zero.

Answer 4

This has been answered with an explanation. Why are you asking it again?

Limit is 0. See previous comments.