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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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    $\begingroup$ Usually people would do that with L'Hospital, byt writing $\ln x/ (1/x)$. $\endgroup$
    – Julien
    Feb 14, 2013 at 13:42
  • $\begingroup$ You want to take the limit from the right... $\endgroup$ Feb 14, 2013 at 13:42
  • $\begingroup$ You can put $n=1/x$ in this other new question: math.stackexchange.com/questions/303933 $\endgroup$
    – GEdgar
    Feb 14, 2013 at 14:02
  • $\begingroup$ A wrong answer : $x\ln x=x+\int_0^x \ln u\,du$ so has limit $0$ when $x\to 0$. :p $\endgroup$
    – zwim
    Jan 6, 2017 at 8:07

5 Answers 5

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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 polynomial.

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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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Putting $y = 1/x$, $x^x =(1/y)^{1/y} =\dfrac1{y^{1/y}} $.

So, if we can show that $\lim_{y \to \infty} y^{1/y} = 1$, then $\lim_{x \to 0} x^x = 1$.

Here (again) is a simple proof that $n^{1/n} \to 1$ as $n \to \infty$.

By Bernoulli's inequality, $(1+n^{-1/2})^n \gt 1+n^{-1/2}n =1+n^{1/2} > n^{1/2}$.

Raising to the $2/n$ power, $(1+n^{-1/2})^2 > n^{1/n}$ so that $n^{1/n} < (1+n^{-1/2})^2 = 1+2n^{-1/2} + n^{-1} < 1+3n^{-1/2} $.

Since $n^{1/n} > 1$, $n^{1/n} \to 1$.

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$(x\ln x)'=\ln x+1\to -\infty$ when $x\to 0^+$ we conclude that $(x\ln x)$ is a decreasing function near $0$.

Since $x\ln(x)\le 0$ then for $x$ sufficiently small we have $|x\ln x|<C$ for some constant $C$.

Let's have $x\to 0^+$, $y\to 0^+$ so that $u=xy\to 0^+$.

$|u\ln u|=|xy\ln(xy)|=|xy(\ln x+\ln y)|=|y(x\ln x)+x(y\ln y)|\le|y(x\ln x)|+|x(y\ln y)|\le C(|x|+|y|)\to 0$

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