Rational expression limit problem

StonedPanda
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Sligtly more complex than the average one, I'd assume. How would I go about proving that the limit of of a rational expression consisting of two polynomials of the same degree goes to one and the limit of one where the degree of the bottom is greater than the degree of the top goes to zero. I'd imagine i'd have to use induction, but I've never learned it!

This step is actually a step in a more complicated problem, which I've got for the most part.
 
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Limit as x approaches infinity? Consider:

2x/x

this is a rational function of two polynomials of the same degree whose limit is never 1 for any x, it is always 2. Also consider:

2(x+1)/x

again, it has a limit of 2 as x approaches infinity, but it's limit isn't always 2, for example, it is 4 when x approaches 1. Perhaps you mean when both polynomials also have the same leading coefficient. In that case, the limit as x approaches infinity will be 1.
 
yes, sorry, i meant for the case when the polynomials has the same leading coefficient as the variable approaches infinity.
 
so can anyone help me prove this? I'm trying to teach myself induction...
 
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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