Can anybody explain this to me?

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In this link:


http://people.ischool.berkeley.edu/~johnsonb/Welcome_files/104/104hw9sum06.pdf

For number 28.8 b),

...for case 1, they say that x is the limit for the sequnce <x_n>, right? So doesn't the limit for the sequence <f(x_n)> need to be x^2? Why does the answer say that it must be zero?
 
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That would be true for f continuous, which the exercise is proving to be false. That approach seems a bit complicated.
if x_n is always rational
lim <f(x_n)>=0
if x_n is never rational
lim <f(x_n)>=x^2

they agree if and only if x=0
 
Hi Artusartos! :smile:
Artusartos said:
For number 28.8 b),

...for case 1, they say that x is the limit for the sequnce <x_n>, right? So doesn't the limit for the sequence <f(x_n)> need to be x^2? Why does the answer say that it must be zero?

No, it says limn->∞xn2 = x2 0. :wink:

(not equal to)
 
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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