Limits

Calculus Level 3

Evaluate $\displaystyle \lim_{x\rightarrow 0^+} x^x$ .

Hint: $x^x = e^{x\ln x}$ .

The answer is 1.

1 solution

Zach Abueg
Jan 19, 2017

$\lim_{x \to\ 0} x^x = 0^0$ , which is indeterminate.

To overcome this, we are given a hint:

$x^x = e^{x lnx}$

Knowing this, let's evaluate $e^{\lim_{x \to\ 0} x lnx}:$

$e^{\lim_{x \to\ 0} 0 \times ln (0)} = e^0 = 1$

$0 \times \ln(0)$ is of the indeterminate form $0 \times -\infty$ , so we need to use L'Hopital's rule to evaluate

$\displaystyle\lim_{x \to 0} (x \times \ln(x)) = \lim_{x \to 0} \dfrac{\ln(x)}{\dfrac{1}{x}} = \lim_{x \to 0} \dfrac{\dfrac{1}{x}}{-\dfrac{1}{x^{2}}} = \lim_{x \to 0} (-x) = 0$ .

Brian Charlesworth - 4 years, 4 months ago

Ah yes, I was thinking it was $1$ at the time for some reason. Thank you!

Zach Abueg - 4 years, 4 months ago

Limits

The answer is 1.

1 solution

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