Cross Product Continuity: Showing Definition is Satisfied

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1. Show that the cross product is a continuous function

3.
I have tried to apply the definition of continuity: find a delta such that
|x-y|< delta implies |f(x)-f(y)|< epsilon
but I'm having trouble finding a delta that would take me to the conclusion.
 
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Perhaps you should show us what you've tried. What is |f(x)-f(y)| in this case?
 
In this case, |f(x)-f(y)| is a real number (the length of a vector in R^3).
A better question is what is |x-y|. x is one pair of vectors in R^3 and y is another pair of vectors in R^3. Then what is |x-y|?
I have tried relating |x-y| to the fact that |a x b|=|a||b|sin(theta) but can't get anything that seems to be on the right direction.
 
I figured it out.
 
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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