# Example of a limit with indeterminate form $0^0$ equal to $1/\pi$ [closed]

I'm working on a calculus problem and I'm trying to come up with a function that has an indeterminate limit of the form $$0^0$$, but which actually evaluates to a non-zero value of $$\frac{1}{\pi}$$. I'm having trouble coming up with an example, so I was hoping someone here might be able to help me out.

To be more specific, I'm looking for two functions $$f(x)$$ and $$g(x)$$ such that:

$$\lim_{x\to0} f(x)^{g(x)} = 0^0$$ (an indeterminate form), but in fact, the limit is equal to $$\frac{1}{\pi}$$.

If anyone has any suggestions or ideas for functions that could meet these requirements, I would really appreciate it. Thanks in advance!

• "Such that $\lim_{x\too0} f(x)^{g(x)}=0^0$" is an abuse of notation. You can just write $f(x)^{g(x)}$ is in the form $0^0$ near $x=0.$ Commented Feb 21, 2023 at 17:01
• Let $f(x)=(1/\pi)^{1/x^2}$ and let $g(x)=x^2$. Commented Feb 21, 2023 at 19:42

Take $$f(x)=\dfrac{x}{\sqrt[x] \pi}$$ and $$g(x)=x$$, you have: $$\lim_{x\to 0^+}\left(\frac{x}{\sqrt[x] \pi}\right)^x=\frac{1}{\pi}$$
• We don't know the purpose of the OP. But a more satisfying answer might be one that does not explicitly involve $\pi$. Commented Feb 21, 2023 at 18:20
Take $$f(x):= \left(2\tan^{-1}\left(\frac{1}{x^{2}}\right)\right)^{-1/x^2},g(x):=x^2.$$
Then $$\lim_{x\to 0} f(x)^{g(x)}=\dfrac{1}{\pi}$$