The tutor works lim(x→0+) x^x.

The ability to evaluate a limit often depends on what’s in your “bag of tricks.” Here’s a first-year limit that uses a few:

Example:

Find limit_x→0+x^x

Solution:

We can imagine

y=limit_x→0+x^x

then use the log trick

lny=ln(limit_x→0+x^x)

Provided everything stays defined (in this case, x>0),

lny=limit_x→0+lnx^x (important!)

Now, we can simplify the inside right using the exponent-to-multiple-rule :

lny=limit_x→0+xlnx

We can rewrite as

lny=limit_x→0+lnx/(1/x)

which is the indeterminate form -∞/∞ and therefore eligible for l’Hôpital’s rule.

We proceed by taking separate derivatives top and bottom:

lny=limit_x→0+(1/x)/(-1/x^2)=limit_x→0+-x=0

Recall from the beginning that y=limit_x→0+x^x. Since e^y is the inverse of lny,

lny=0 leads to

y=e^0=1

Therefore

limit_x→0+ x^x = 1

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus, Part One, 3rd Edition. Toronto:   D.C. Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.