How do I show that x^x->1 as x->0?

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How do I show that x^x->1 as x->0?
 
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Try using log's to convert the exponentiation into a product.
 
I can't find a lower bound for xlog(x).
 
use l'hopital's rule.

if you have lim x log x then x log x is the same as [log x / (1 / x)]
 
Got it. Thanks.
 
You can also change it to what may be a more familiar limit by setting x=1/y.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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