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I was thinking about the reasons behind $0^0=1$ and I remember one of my friends studying math arguing about the continuity of the function $x^x$ in $0$. But when I write as $$x^x=e^{x\ln x}$$ I am now looking at $$\lim_{x\rightarrow 0} x\ln x$$ Graphically I can see in Mathematica that it goes to $0.$ But I can't calculate by using a Taylor expansion, because I can't expand log around $0$. How do you prove that?

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Usually people would do that with L'Hospital, byt writing $\ln x/ (1/x)$. –  1015 Feb 14 '13 at 13:42
    
You want to take the limit from the right... –  David Mitra Feb 14 '13 at 13:42
    
You can put $n=1/x$ in this other new question: math.stackexchange.com/questions/303933 –  GEdgar Feb 14 '13 at 14:02

3 Answers 3

up vote 3 down vote accepted

Substitute $x = e^y$, then if $x \to 0$, $y\to -\infty$. So the limit becomes $$\lim\limits_{x\to 0} x \ln x = \lim\limits_{y\to -\infty} y\cdot e^{y} = 0 $$ , because $e^\cdot$ is stronger than any polynominal.

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L Hospitals rule can be used,

$$\lim_{x\to 0}x\ln x=\lim_{x\to 0}\frac{\ln x}{\frac{1}{x}}$$

Using L hospitals Rule (as it has$\frac{-\infty}{\infty } $ form)we have,

$$\lim_{x\to 0}\frac{\ln x}{\frac{1}{x}}=\lim_{x\to 0}\frac{\frac{d\ln x}{dx}}{\frac{d\frac{1}{x}}{dx}}=\lim_{x\to 0}\frac{\frac{1}{x}}{\frac{-1}{x^2}}=\lim_{x\to 0}(-x)=0$$

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$ln(x) = -\sum_{k=1}^\infty {(-1)^k (x-1)^k \over k}$ for $|x-1|<1$

Then $xln(x) = -\sum_{k=1}^\infty {x(-1)^k (x-1)^k \over k}$ wich converges for $x=0$.

q.e.d.

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