Calculus II Limit

Question

Find lim(cosx)^csc^2x

x->0

Answer 1

ln y = (cscx)^2 ln cosx = ln cosx /. (sinx)^2

it is now indeterminant 0/0

Lhopitals rule: -tan / 2 sin cos <--- still 0/0

Lhopitals again: -sec^2 / ( 2cos^2 - 2sin^2)

converges to -1/2

Answer 2

Hi David:

Let's see if we can help here!

lim cos x^{(cosecx)^2} We can't use L'Hopital's Rule because this -> 1^∞ which isn't one of the forms 0/0, etc.

x->0

Suppose the limit = L, if it exists.

Then, L = lim cos x^{(cosecx)^2} => log L = log (lim cos x^{(cosecx)^2})

x->0 x->0

= lim log (cos x^{(cosecx)^2}) since we can take the log inside the lim

x->0

= lim cosec²x log (cos x) by the law of logs

x->0

= lim log (cos x) / sin²x which is of form 0/0 at x = 0

x->0

= lim (1 / cos x) (-sin x) / 2 sin x cos x by L'Hopital's Rule

x->0

= lim -1/2 cos²x

x->0

= -1/2 since cos 0 = 1

∴ log L = -1/2

=> L = e^-1/2

= 1 / √e