Indeterminate form: $0^0$ It is known that $\lim_{x\to 0^+}x^x=1$,
$\lim_{x\to 0^+}0^x=0$ and $\lim_{x\to 0}x^0=1$. So sometimes $0^0$ is left undefined, sometimes defined as $1$.
A question then come to my mind:
Given $(s_n)_{n\in\mathbb N},(t_n)_{n\in\mathbb N}$ are sequences$,\forall n\in\mathbb N,(s_n)\gt0\land(t_n)\gt0$, $\lim_n s_n=\lim_n t_n=0$.
Given $a\in[0,1]$,When will $(s_n^{t_n})\to a$? Even giving more example may help.
I have found some example,
1)$\forall n\in\mathbb N, s_n=t_n=\frac{1}{n},(s_n^{t_n})\to 1$
2)$\forall a\in(0,1), \forall n\in\mathbb N, s_n=a^n\land t_n=\frac{1}{n},(s_n^{t_n})\to a$
3)$\forall n\in\mathbb N, s_n=\frac{1}{n^n}\land t_n=\frac{1}{n},(s_n^{t_n})\to 0$
I think it sounds very 'easy' for $(s_n^{t_n})\to 1$, but are there similarity between the examples?
Or Can I even ask for:
If $\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}g(x)=0, a\in\mathbb R$, when will $\lim_{x\to 0^+}f(x)^{g(x)}=a$?
Thanks.

Note that we need $f(x)\to 0^+$ for the expression to be well defined, then
$$f(x)^{g(x)}=e^{g(x)\log(f(x))}\to a>0$$
since all depends upon
$$\lim_{x\to 0^+} g(x)\log(f(x))$$
For the case $a=+\infty$ let consider


*

*$f(x)=x \to 0$

*$g(x)=-|\log x|^{-\frac12}\to0$


indeed since $x\to 0^+$
$$g(x)\log(f(x))=-|\log x|^{-\frac12}\log x=-\frac{\log x}{|\log x|^\frac12}=\frac{|\log x|}{|\log x|^\frac12}=|\log x|^\frac12\to +\infty$$
