Does the following formula $$\lim f(x)^{g(x）}=(\lim f(x))^{\lim g(x)}$$
always hold?
If not, why?
And what about $\lim f(x)\cdot g(x)$? Is that the same as $\lim f(x)\cdot \lim g(x)$?
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3$\lim_{n\to +\infty}\left(1+\frac{1}{n}\right)^n = e \neq 1^\infty$ – Jack D'Aurizio Oct 22 '18 at 12:20
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For questions like this, see "indeterminate forms" in your calculus textbook, or https://en.wikipedia.org/wiki/Indeterminate_form – GEdgar Oct 22 '18 at 12:45
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Could you clarify the questio you added "...and what about$ \lim f(x)*g(x)$?Is that the same as $\lim f(x)^{g(x)}$?" – user Oct 22 '18 at 13:01
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I mean is that lim f(x)g(x)=lim f(x)lim g(x) always holds? – 占俊涵 Oct 22 '18 at 13:05
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@ GEdgar thanks – 占俊涵 Oct 22 '18 at 13:11
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@占俊涵 Yes, that's a simpler case and the equality holds always when $\lim f(x)=L_1\in R$ $\lim g(x)=L_2\in R$ since $L_1\cdot L_2$ is always well defined. – user Oct 22 '18 at 13:14
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I would say: limit yourself to the case $f(x)>0$ and define $$f(x)^{g(x)} = \exp\big(g(x)\log f(x)\big).$$ Then you reduce to cases involving continuity of the product and the function $e^x$.
GEdgar
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That method is fine and the best one to evaluate $\lim f(x)^{g(x)}$ but in that case it seems to me that the OP is asking a slightly different thing that is when can we apply $\lim f(x)^{g(x）}=(\lim f(x))^{\lim g(x)}$ which, as you also noticed commenting my answer, can include cases with $f(x)\le 0$. – user Oct 22 '18 at 12:36
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+1 This is way to go because the symbol $a^b$ is most easily handled by writing it as $\exp(b\log a) $ and the problem thus shifts to the limit of $g(x) \log f(x) $. – Paramanand Singh Oct 23 '18 at 04:01