Set X is defined with $$\sqrt {(x (x + 1))}/(2 x + 1)$$, x > 0. I can't prove that for every epsilon greater than zero there exists an t > 0 such that $$1/2 - \sqrt{(t (t + 1))}/(2 t + 1) < \epsilon.$$ Every other step I get.
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It is $$\frac{\sqrt{x(x+1)}}{2x+1}\le \frac{1}{2}$$ for $x>0$, this is true, multiplying by $2$ and squaring we get $$4x(x+1)\le 4x^2+4x+1$$ This is true since we get $$0\le 1$$
Dr. Sonnhard Graubner
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I know that it is greater than zero, but how can I squeeze it between zero and epsilon? Can't I just then replace 1/2 with 1 and still get a true statement? – user680838 Oct 24 '19 at 11:32
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Sorry those are two seperate questions, in the second one I mean that if I replace 1/2 with 1 wouldn't I by definition also get that supX = 1 instead of 1/2? – user680838 Oct 24 '19 at 11:40