Calculating Limit: How to Show Lim x^2-sin(x)→∞

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How do I formally show that lim x^2 - sin(x) as x tends to infinity is infinity?
 
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I suppose you could look at the minimum possible value of your function... since sin(x) can never be greater than one, your equation can never be less than x^2-1. So work with that instead.
 
There's a theorem that is sometimes called the Squeeze Theorem or Squeeze Play Theorem. If f(x) <= g(x) <= h(x) for all x in a suitable domain, and lim f(x) = lim h(x) = L, then lim g(x) = L.

x^2 - 1 <= x^2 - sin(x) <= x^2 + 1 for all x. What can you say about lim x^2 -1 and lim x^2 + 1 as x grows large without bound?
 
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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