Curious about expressions like \( x^x \)? In this post, I’ll explain how you can use L’Hospital’s Rule in unexpected ways to solve this problem:
$$ \lim_{x \to 0^+} x^x = 1 $$
Note: \( x \to 0^+ \) signifies that \( x \) approaches \( 0 \) from the right.
It turns out that the above problem is related to my previous post, which describes the following problem:
$$ \lim_{x \to 0^+} x \ln(x) = 0 $$
To see how that is so, consider \( x^x \). In general,
$$ x = e^{\ln(x)} $$
That is because the exponential function and \( \ln(x) \) cancel each other out.
So, I can write:
$$x^x = (e^{\ln(x)})^x $$
It turns out that
$$ (e^{\ln(x)})^x = e^{x \ln(x)}. $$
So,
$$ x^x = (e^{\ln(x)})^x = e^{x \ln(x)}. $$
Now, we can solve the problem. Since \( x^x = e^{x \ln(x)} \),
$$ \lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x \ln(x)}. $$
It turns out that
$$ \lim_{x \to 0^+} e^{x \ln(x)} = e^A $$
where
$$A = \lim_{x \to 0^+} x \ln(x). $$
This is where L’Hospital’s Rule comes in. As explained in my previous post, this rule can be used to calculate the following:
$$ \lim_{x \to 0^+} x \ln(x) = 0 $$
So, \( A = 0 \). Recall that any number to the power of zero is one. So as
$$ \lim_{x \to 0^+} e^{x \ln(x)} = e^A $$
We have
$$ \lim_{x \to 0^+} e^{x \ln(x)} = e^0 = 1 \,. $$
So as
$$ \lim_{x \to 0^+} x^x = \lim_{x \to 0^+} e^{x \ln(x)}, $$
$$ \lim_{x \to 0^+} x^x = 1 $$
which is the answer we’re looking for.
