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Suppose that we have $\lim_{x\to a}f(x)^{g(x)}$ with $f(a)=0$ and $g(a)=0$.

Can we say, that the value of the above limit is $0^0=1$?

Is there a counterexample?

UPDATE: I already saw examples you've provided. Is there a counterexample with continuous functions?

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    You should rule out $f(x)\equiv 0$ as a trivial counterexample. – Peter Jan 23 '22 at 14:27
  • May I ask you to make it a bit clear? Thanks – Mikasa Jan 23 '22 at 14:29
  • This is an indeterminate form, so it can do just about anything. – Randall Jan 23 '22 at 14:30
  • Is there no hypothesis about $f$ and $g$? Then consider $f $ defined by $f(x) = 2$ for $x\ne a$ and $f(a)=0$, and $g=f$. Then $\lim_{x\to a} f(x)^{g(x)} = 2^2 = 4$. – jjagmath Jan 23 '22 at 14:31
  • Even if we have $f(x)=g(x)=x$ and $a=0$ , the conditions are satisfied, but the limit does not even exist (only the limit from the right exists) – Peter Jan 23 '22 at 14:41

1 Answers1

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Counter-example: $\lim_{x\to 0} $[1-cosx$]^x$ ,where [ ] stands for greatest integer function.

Edit: Another example I found in the below linked answer : $\lim_{x\to 0+}x^\frac{1}{\ln(x)}$

A similar discussion which took place here: Zero to the zero power – is $0^0=1$?