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Can anyone clear me this fundamental concept about this. I am confused for many months over this.

It is said that in $\lim_{x\to a} f(x)^{g(x)}$.
$f(x)$ should be greater then $0$.
Can anyone explain the reason?

If $f(x)$ is $-2$ and $g(x)$ is $\frac{1}{2}$,then it is not possible. But if $f(x)$ is $-2$ and $g(x)$ is $2$,then why it is not possible? (It is true as $(-2)^2 = 4$)

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As it has been pointed out, $\lim_{x \to a} (-2)^{1/2} = i\sqrt{2}$ so using complex numbers changes this. But if you haven't learned about complex numbers, it often said that $f(x)$ in $\lim_{x\to a}f(x)^{g(x)}$ has to be positive, mostly because it is easier to do that way. As you have shown there are cases where the limit exists even though $f(x)$ isn't positive, but they are few, and uncommon enough (at least until you have reached a level where you will have learned about complex numbers) that they can dealt with.

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If you are doing real analysis, this proviso means that you don't run into any problems - the exponential function can be defined.

As others have noted in threads linked in the comments, it is possible to extend the definition of the exponential function, but there are some technicalities involved which impinge on whether it makes sense to take a limit at all - whether the function you define is continuous, for example.

And don't forget that in taking the limit you have specified $g(x)$ varies as well as $f(x)$ so you are potentially dealing with a continuously changing real exponent rather than some nice integer or rational one - so you really need the exponential function to be valid for all reals, and continuous in the exponent. If $f(x)$ is positive, there isn't a problem.

This doesn't mean it can't be done for other values, of course. But there are pitfalls for the unwary.

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