Limit of a Sequence: Does Square or Sqrt Change It?

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Homework Statement
I was curious, if I have a sequence that has a limit of 1
Relevant Equations
Lim an=1 as n tends to inf
Lim of an^2=1 as n tends to inf
Does the square of the sequence also have a limit of 1. Does the square root also equal 1? I've been trying to find some counterexamples but I think the limit doesn't change under these operations?
 
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Karl Porter said:
Homework Statement:: I was curious, if I have a sequence that has a limit of 1
Relevant Equations:: Lim an=1 as n tends to inf
Lim of an^2=1 as n tends to inf

Does the square of the sequence also have a limit of 1. Does the square root also equal 1? I've been trying to find some counterexamples but I think the limit doesn't change under these operations?
Can you think of a basic theorem of limits that would lead to a one line proof?

Hint: If ##\lim_{n \rightarrow \infty} a_n = L_a## and ##\lim_{n \rightarrow \infty} b_n = L_b##, then ...
 
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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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