Question

Lim x-->π/2 (cos x)^cos x?

Answer 1

`lim_(x->pi/2) (cosx)^(cosx) = 1`

Explanation:

Given the function: `f(x) = (cosx)^(cosx)` consider:

`g(x) = ln (f (x)) = ln ((cosx)^(cosx)) = cosx ln cosx`

substitute `y= cosx` so that:

`lim_(x->pi/2) cosx ln cosx = lim_(y->0) ylny`

This limit is in the form `0 xx -oo`, but we can transform it in the form `oo/oo` and then apply l'Hospital's rule:

`lim_(y->0) ylny = lim_(y->0) lny/(1/y) = lim_(y->0) (d/dy lny)/(d/dy 1/y) = lim_(y->0) (1/y)/(-1/y^2) = lim_(y->0) -y = 0`

Then we know that:

`lim_(x->pi/2) g(x) = 0`

As `e^x` is a continuous function for all `x in RR` we also have that:

`lim_(x->pi/2) e^(g(x)) = e^(lim_(x->pi/2) g(x) ) =e^0=1`

But `e^(g(x)) = e^(ln(f(x))) = f(x)` so we can conclude that:

`lim_(x->pi/2) (cosx)^(cosx) = 1`