Convergence of orthagonal vectors

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Homework Statement


{uk} is in Rn and converges to u in Rn

let v be in Rn and v is orthogonal to each uk.

prove v is orthogonal to u


Homework Equations


just definition of convergence. and orthogonality. <v,u>=0 if v is orthogonal to u.

The Attempt at a Solution


uk converges so it is cauchy, so it's terms are getting closer to each other.

for epsilon>0 , there exists k>= k0 st. ||uk-u|| < epsilon
so if v is orthogonal to uk then u is orthogonal to each term in uk. but the terms of uk are getting closer to u. so if v is orthogonal to a uk that is very close to u, then it is also orthogonal to u.

this proof is in no way formal but i think i have the right idea. can some one please help rewrite this?
 
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It sounds like you want to argue that if uk converges to u, then <uk, v> converges to <u, v>. Does that sound right?

if so, let h be an arbitrarily small vector. what happens to <u + h, v> = <u, v> + <h, v> as h goes to zero? The magnitude can vary of course, But you just want an upper and lower bound on it anyways. Each magnitude of h represents a neighborhood of u that contains every element of uk for k sufficiently large. (this of course comes from the definition of a limit)

Alternatively, you might be able to do something cool with the fact that <u - uk, v> = <u, v>. That that is useful is just a guess on my part though.
 
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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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