Lim x->0+ | ln x|^sinx using l'hospital's?

Answer 1

[x→0⁺]lim |ln x|^sin x

First, we resolve the absolute value. When x is near 0, ln x is always negative, so this is:

[x→0⁺]lim (-ln x)^sin x

Now, as is typically the case where the indeterminate form involves exponents, it is easier to evaluate the logarithm of the limit than the limit itself. So we want to find:

ln [x→0⁺]lim (-ln x)^sin x

Since the logarithm is continuous, we may move it inside the limit:

[x→0⁺]lim ln ((-ln x)^sin x)

Using the laws of logarithms:

[x→0⁺]lim sin x * ln (-ln x)

Now, this is of the form 0*∞. We transform it into ∞/∞ form thus:

[x→0⁺]lim ln (-ln x)/csc x

Now, we employ L'hopital's rule:

[x→0⁺]lim (1/(-ln x) * -1/x)/(-csc x cot x)

Rearranging:

[x→0⁺]lim -(sin x tan x)/(x ln x)

Now, we may split this up into the product of two limits (both of which we will show exist):

[x→0⁺]lim sin x/x * [x→0⁺]lim -tan x/ln x

The first limit is of course 1, and the second takes the form 0/∞, which is 0. So we have that:

ln [x→0⁺]lim (-ln x)^sin x = 0

Of course, we are interested in [x→0⁺]lim (-ln x)^sin x itself, not its log, so exponentiate both sides to obtain:

[x→0⁺]lim (-ln x)^sin x = 1