Limit Of A Rapidly Increasing Exponential

Calculus Level 1

$\large \displaystyle \lim_{x\to 0^+} x^x =\, ?$

$\begin{aligned} 0.5^{0.5} &=& 0.7071\ldots \\ 0.4^{0.4} &=& 0.6931\ldots \\ 0.3^{0.3} &=& 0.6968\ldots \\ 0.2^{0.2} &=& 0.7247\ldots \\ 0.1^{0.1} &=& 0.7943\ldots \\ \end{aligned}$

0

1

\frac{e}{3}

\frac{\sqrt{2}}{2}

Limit does not exist

2 solutions

Brandon Stocks
Apr 27, 2016

Relevant wiki: Limits by Logarithms

$x^{x} = e^{\ln(x^{x})} = e^{x \times \ln(x)}$

Consider the exponent by itself

$x \times \ln(x) = \frac{\ln(x)}{x^{-1}}$

Note that as positive $x$ goes to zero both the numerator and denominator go to infinity, so L'Hospital's Rule can be applied.

$\frac{d}{dx}$ $\ln(x) = x^{-1}$ and $\frac{d}{dx}$ $x^{-1} = -x^{-2}$

$\displaystyle \lim_{x \to 0+} \frac{\ln(x)}{x^{-1}} = \lim_{x \to 0+} \frac{x^{-1}}{-x^{-2}} = \lim_{x \to 0+} -x = 0$

Thus $\displaystyle \lim_{x \to 0+} x^{x} = e^{0} = 1$

Moderator note:

Good clear explanation.

Note that instead of saying "consider the exponent by itself", what we are doing is applying the following theorem:

If $\lim_{x \rightarrow a } f(x)$ exists and is equal to $L$ (finite), and $g(x)$ is a continuous function at $x = a$ , then $\lim_{x \rightarrow a } g\circ f(x) = g(L)$ .

Divyansh Verma
Nov 27, 2017

Lim x→0 x^x = Lim x→0 x^(Lim x→0 x)= x^0=1

Limit Of A Rapidly Increasing Exponential

2 solutions

Moderator note:

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