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How do you find the limit of #x^x# as #x>0^-#?

Answer 1

Christopher Allers

#lim_(x->0^+) x^x = 1#

Write the function as:

#x^x = (e^lnx)^x =e^(xlnx)#

Note now that:

#lim_(x->0^+) xlnx#

is in the indeterminate form #0*oo#. We can reconduce it to the form #oo/oo# and then apply l'Hospital's rule:

#lim_(x->0^+) xlnx = lim_(x->0^+) lnx/(1/x) = lim_(x->0^+) (d/dx lnx)/(d/dx 1/x) = lim_(x->0^+) (1/x)/(-1/x^2) = lim_(x->0^+) -x = 0#

As #e^x# is a continuous function near #0#:

#lim_(x->0^+) e^(xlnx) = e^(lim_(x->0^+) xlnx) = e^0 = 1#

graph{x^x [-10, 10, -5, 5]}

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Answer 2

Charles Arocha

To find the limit of x^x as x approaches 0 from the negative side, we can use the concept of L'Hôpital's rule. By taking the natural logarithm of both sides, we can rewrite the expression as ln(x^x). Applying the power rule of logarithms, this becomes x ln(x). Now, we can differentiate both the numerator and denominator with respect to x. The derivative of x is 1, and the derivative of ln(x) is 1/x. Applying L'Hôpital's rule, we get the limit as x approaches 0 from the negative side to be 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #x^x# as #x&gt;0^-#?

How do you find the limit of #x^x# as #x>0^-#?