What is the solution to $\lim_{x\to0^-} x^x?$

Question

$$\lim_{x\to0^-} x^x.$$

Symbolab says it's 1. My answer sheet says it does not exist. Which is correct, and can you explain the solution?

Answer 1

The term $$x^x$$ is only defined for all reals if $$x>0$$ holds.

Answer 2

Is it considered over real domain? If so, then the limit does not exists, because $$x^x=e^{xlog(x)}$$ and one cannot take a logarithm of a non-positive number.

Answer 3

EDIT: I see the question has changed slightly since I began typing the first draft of this. I added clarification to answer the actual question, and left the rest of the post in its entirety in case it helps a future viewer.

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$\lim_{x \to 0} x^x$ does not exist.

$\lim_{x \to 0^+} x^x = 1$.

Note the subtle difference. In the limit that exists, we approach $x$ only from the right. The limit that doesn't exist is the two-sided limit. This is because $x^x$ is not defined for $x < 0$, therefore we cannot approach $x=0$ from the left. Note that this also implies that $\lim_{x \to 0^-} x^x$ does not exist.

As for why the defined limit is 1:

$$x^x = e^{\ln x^x} = e^{x \ln x}$$

Therefore,

$$\lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x \ln x} = e^{\lim_{x \to 0^+} x\ln x}.$$

We can push the limit inside the exponent like that because limits can be pushed into and pulled out of continuous functions.

Anyway, we can use l'Hôpital to evaluate $\lim_{x \to 0^+} x\ln x$.

\begin{align} \lim_{x \to 0^+} x \ln x &= \lim_{x \to 0^+} \frac{\ln x}{1/x}\\[0.3cm] &= \lim_{x \to 0^+} \frac{1/x}{-1/x^2}\\[0.3cm] &= \lim_{x \to 0^+} (-x)\\[0.3cm] &= 0 \end{align}

Therefore $\lim_{x \to 0^+} x^x = e^0 = 1$.

Answer 4

The problem is that $x^x$ is ill-defined when $x < 0$, which is the domain your limit is considering. For example, $(-0.5)^{-0.5}$ is not defined. $(-1/3)^{-1/3}$ exists, but $(-1/4)^{-1/4}$ does not, and so on. In fact there are negative values of $x$ arbitrarily close to $0$ so that $x^x$ is not defined. By the definition of the limit, in such a case the limit cannot be defined.