find the limit as x aproaches pi/2 from the left for cosx^cosx?

Answer 1

[x→π/2⁻]lim (cos x)^(cos x)

This is a 0^0 form, so it cannot be evaluated directly. So instead, write it in exponential form, and try to find the limit of its logarithm.

[x→π/2⁻]lim e^(cos x ln (cos x))
e^([x→π/2⁻]lim cos x ln (cos x))
e^([x→π/2⁻]lim ln (cos x)/sec x)

The limit is now a -∞/∞ form, so using L'hopital's rule:

e^([x→π/2⁻]lim -tan x/(sec x tan x))
e^([x→π/2⁻]lim -1/sec x)
e^([x→π/2⁻]lim -cos x)
e^0
1

And we are done.

Edit: Fixed a dropped negative sign.

Answer 2

y = cosx^cosx
lny = cosx ln(cosx)
= ln(cosx) / [1/cosx]
as x ->pi, lny -> -infinity/ infinity
so L'Hospital rule applies
Let u = cosx which ->0 as x -> pi/2
lny = ln(u) / (i/u) take limit as u -> 0
= f(x)/g(x) limit = f'(0)/g'(0)
= 1/u / (-1/u^2) = -u ->0 as u ->0
so
lny -> 0 as x -> pi/2, or y -> 1 (ln(1) = 0)
or y = cosx^cosx -> 1 as x -> pi/2