Need help in evaluating an indeterminate form ($0^0$) of a limit

Question

Let, $\Psi(x)= f(x)^{g(x)}$

Given, $\lim_\limits{x\to a} f(x)= 0, \lim_\limits{x\to a} g(x)=0$ and $\lim_\limits{x\to a} \Psi(x)$ exists.

If the mentioned conditions are guaranteed to be fulfilled, is it always true that $$\lim_{x\to a} \Psi(x) =1 $$

Answer 1

No.

In fact, the limit can be anything. For example, if you want the limit to be $c\ne 1$: Let $a=0$, $f(x) = e^{-1/|x|}$, and $g(x) = (-\log c)|x|$, then $\Psi(x) = f(x)^{g(x)} = c$ for all $x\ne 0$, but $$\lim_{x\rightarrow 0} f(x) = \lim_{x\rightarrow 0} g(x) = 0$$

If $f$ and $g$ are analytic functions at $0$ then the answer is yes! In that case, one can write $f(x) = x^m \hat f(x)$ and $g(x) = x^n \hat g(x)$ for some positive integers $n,m$ and analytic functions $\hat f,\hat g$ that are nonzero at $0$. (In order for the limit to exist, you also need $\hat g(0) >0 $, but you did mention that a constraint in the question). In that case:$$ \Psi(x) = f(x)^{g(x)} = x^{m x^n g(x)} \hat f(x)^{x^n \hat g(x)} = (x^{x^n})^{m \hat g(x)} (\hat f(x)^{\hat g(x)})^{x^n} $$ By continuity, you can see that the limit of $(\hat f(x)^{\hat g(x)})^{x^n}$ as $x$ goes to $0$ must be $1$. Also, because the limit of $(x^{x^n})$ is $1$, and $m\hat g(x)$ is bounded in a neighborhood of $0$, you will also have the limit of the first term $(x^{x^n})^{m \hat g(x)}$ is $1$.

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In general, the easiest way to evaluate limits of the form $0^0$ is by taking a logarithm:$$ \log (f(x)^{g(x)}) = g(x)\log f(x)\rightarrow 0\cdot(-\infty) $$ and $0\cdot\infty$ is much easier to handle (e.g., change it to $\frac00$ or $\frac\infty\infty$ and use L'Hopital's rule).

Answer 2

No

If $f(x)=0$ and $g(x)$ any strictly positive function which converges to zero. Then $\Psi = 0$

Answer 3

No. See for instance $f(x)=e^{-(x-a)^{-2}}$ and $g(x)=-(x-a)^2\ln\lvert x-a\rvert$.

Answer 4

Write the function as $$e^{g\log f}.$$ Then that's the same as determining $g\log f$ when both $f$ and $g$ vanish. But then $\log f=-\infty.$ So write it as $$\frac{\log f}{1/g}=\frac{-\infty}{\pm\infty}.$$ Clearly this can take any value at all, since there are infinitely many infinites. What obtains is given by the form of $1/g.$ For example, if $1/g=f,$ then the limit is $0.$ If $1/g=\log f,$ clearly it is $1.$ If $1/g=\sqrt {\log f},$ the limit is infinite, and so on.