Limit Comparison Theorem: Proving the Relationship Between Two Functions

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Homework Statement


Prove that if f(x)<=g(x) then lim f(x) <= lim g(x).

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The Attempt at a Solution



I've tried by definition of limit, but I didn't get anywhere with this... Can anyone help me??
 
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Does the limit of a function ever give you a value that is higher than the range of values of the function itself?
 
Try a "proof by contradiction". Suppose lim f(x)> lim g(x). Let \alpha= lim f(x)- lim g(x) and choose \epsilon= \alpha/2.

I am puzzled by phind's question. The answer is "yes, it does" but I don't see how that helps here.

(Note, by the way, if the condition were "f(x)< g(x)" then it would NOT be true that "lim f(x)< lim g(x)". Phind's suggestion would be helpful in proving that.)
 
HallsofIvy said:
I am puzzled by phind's question. The answer is "yes, it does" but I don't see how that helps here.

QUOTE]

Hm ... I guess I'm missing something. I don't see how the limit of a function could possibly give you a value that is outside the range of the possible values of the function.
 
Hm ... I guess I'm missing something. I don't see how the limit of a function could possibly give you a value that is outside the range of the possible values of the function.

Try ##\lim_{x\to\infty}\frac 1 x##.
 
Last edited:
LCKurtz said:
Try ##\lim_{x\to\infty}\frac 1 x##.

OK, thanks. I got it.
 
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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