How do you find a limit as it aproaches infiniti

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Homework Statement


limit x->infiniti sinx=


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as x-->infinity values of sinx fall between 1 and -1 meaning that as x approaches infinity, sinx doesn't approach a specific value and so the limit doesn't exist
 
I was sorely tempted to say that the limit as "it" approaches infinity must be infinity!

Of course, you mean the limit as x approaches infinity. As rock.freak667 said, look at what happens to sin(x) as x gets larger and larger- no matter how large x is, sin(x) keeps varying between -1 and 1. It does not "approach" and one number and so there is no limit.
 
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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