L'Hospital's Rule for the indeterminate form $0^0$

Question

I'm teaching an introductory real analysis course for the first time this year and one of the textbook problems asks to prove the following:

Let $f$ be differentiable on $(a,b)$ and let $c \in (a,b)$. Suppose that $f$ and $f'$ are nonzero on a deleted neighbourhood of $c$, and $\lim_{x \to c} f(x) = 0$. Prove that \begin{align*} \lim_{x \to c} |f(x)|^{f(x)} = 1. \end{align*}

Since this is a limit of the indeterminate form $0^0$, L'Hospital's rule can be applied to evaluate the limit of the logarithm of the expression, which can be showed straightforwardly to be 0. Hence the limit is $e^0 = 1$.

Anyways, this got me to thinking more generally about the case $\lim_{x \to c} |g(x)|^{f(x)}$ where both $g$ and $f$ go to zero at $c$. Can the limit be something other than 1? I tried but I could not come up with a counterexample. After some consideration, it seems to me that one can prove fairly straightforwardly that a limit of this type must always evaluate to 1. So, a couple of questions:

1. Am I wrong? Is there a counter example?
2. (If I am not wrong), is this fact well known? I have looked in several calculus and analysis textbooks and have not seen it mentioned, though the indeterminate form $0^0$ is often given as a case where L'Hospital's rule can be applied. Many different examples are given and they (seemingly) always work out to 1! (Here is a particularly complicated one: https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/lhopitaldirectory/solution24.html ) It seems curious to me that so many examples are given if the result will always be the same...

Answer 1

The obvious counterexample would be $$\lim_{x\rightarrow 0} 0^{|x|} = 0$$ A less obvious counterexample could be $$\lim_{x\rightarrow 0} (e^{-1/|x|})^{|x|} = e^{-1}$$ I recall having read, that if $f$ and $g$ are analytic in a neighborhood of $c$ and $f(x)> 0$ for $x$ sufficiently close to $c$ (but not equal to $c$). Then, if $\lim_{x\rightarrow c} f(x)= \lim_{x\rightarrow c} g(x) = 0$, we would have that $$\lim_{x\rightarrow c} f(x)^{g(x)} = 1.$$ This might explain why the textbook examples usually evaluate to $1$, since most functions that you encounter in an elementary calculus course would be analytic.

Answer 2

Answering "Can the limit be something other than 1?": $$\left( \frac{1}{n}\right)^{\frac{1}{\ln n}} = {\Large e}^{\frac{1}{\ln n}\cdot \ln \frac{1}{n}} = {\Large e}^{-1} $$ Trick in making from $\boldsymbol{f \cdot \ln g}$ something other, then $0$. Another example taking $f=g\to 0$ gives $-\infty$ in power i.e. $0$ in answer.

Answer 3

$g^f$ case. Try this one $$ g(x) = e^{-1/x^2},\qquad \lim_{x\to 0} g(x) = 0,\qquad g'(x) = \frac{2e^{-1/x^2}}{x^3},\qquad \lim_{x\to 0}g'(x) = 0, \\ f(x) = {x^2},\qquad \lim_{x\to 0} f(x) = 0,\qquad f'(x) = 2x,\quad \lim_{x\to 0}f'(x) = 0 \\ f,f',g,g'\quad\text{are all nonzero on }\mathbb R \setminus \{0\} \\ f,f',g,g'\quad\text{all have removable discontinuities at }0 \\ g(x)^{f(x)} = \left(e^{-1/x^2}\right)^{x^2} = e^{-1} \not\to 1 $$

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There is no counterexample for the $f^f$ case. Simply because $\lim_{x \to 0^+} x^x = 1$.