Please find both right hand & left hand limits using l'hopital rule?

Question

lim cos(x)/(1-sin(x))^(2/3)
x tends to pi/2
thanks
is it possible for this limit to be ever found using l'hopital rule?

Answer 1

Madhukar made a mistake
(1 - sin x)^(-1/3) → ∞ as x → π/2
So it is still an indeterminate form after using L'Hopital's rule.

L'Hopital's rule can not be used in this case. It can't be used because (1 - sin(x))^(2/3) is not differentiable at π/2.

Instead, multiply the numerator and denominator by cos(x)
cos^2(x) / [cos(x) * (1 - sin(x))^(2/3)]
use the trig identity sin^2(x) + cos^2(x) = 1
(1-sin^2(x) / [cos(x) * (1 - sin(x))^(2/3)] =
cancel out a 1-sin^2(x) from the numerator and denominator
1 / [cos(x) * (1 - sin(x))^(1/3)] =

The limit of this as x → π/2 is ∞.

Edit: oops, when I substituted cos^2(x) = 1 - sin^2(x) I forgot the squared for the sin. I fixed it above. I also didn't pay attention to left and right limits being different. 1 - sin(x) is never negative but cos(x) is positive as x approaches pi/2 from the left and negative as it approaches from the right.

So the limit is positive infinity from the left and negative infinity from the right.

Answer 2

Whenever on direct substitution, it takes the form 0/0, L'Hospital's Rule can be used. After using L'Hospital's Rule if again it takes 0/0 form, it can be used again till you get answer. I mean in
lim ( x-> a) f(x) /g(x), if f(a) / g(a) = 0/0, you can use L'Hospital's Rule. Then limit = lim(x -> a) f '(x)/g'(x). Again if f '(a) / g'(a) = 0/0, you can use L'Hospital's Rule and write limit = limit(x -> a) f"(x) / g"(x) till it is solved. Suppose at any stage, it takes the form of a finite number divided by zero, it means that the limit does not exist. Now your problem,

lim cos(x)/(1-sin(x))^(2/3)
Since direct substitution makes it 0/0, using L'Hospital's Rule,
= lim -sin(x) / (2/3)(1 - sinx)^(-1/3)*(-cosx)
Now on putting x = π /2, it takes the form (-1)/0. So the limit does not exist.
As x -> π /2 from the left, lim -> + ∞ and
as x -> π /2 from the right, lim -> - ∞

Answer 3

lim cos(x)/(1-sin(x))^(2/3)= -sinx/ 2/3(1-sinx)^-1/3 (-cosx)
3/2 sinx (1-sinx)^1/3 / -cosx = 0/0
then again
3/2 {sinx 1/3(1-sinx)^-2/3(-cosx) +cosx (1-sinx)^1/3}/-sinx
=0/-1 =0