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I have $a>b>0$ and $z = \frac{a-\sqrt{a^2-b^2}}{b}$

As $b\to a$ we have $z \to 1$ and as $b\to 0$ we have $z\to 0$. Is this sufficient to show that $z\lt 1$? If not how can I do it?


$$\frac{a-\sqrt{a^2-b^2}}{b}\lt\frac{a-\sqrt{(a-b)(a+b)}}{b}\lt\frac{a-\sqrt{(a-b)(a-b)}}{b}\lt1$$

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  • $\begingroup$ The first $<$ should be a $=$. You should evaluate $\sqrt{(a-b)(a-b)}$ and then reduce the fraction. $\endgroup$ – Trevor Norton Jan 12 '16 at 22:10
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Try this:

$$\frac{a-\sqrt{(a-b)(a+b)}}{b} < \frac{a-\sqrt{(a-b)(a-b)}}{b} $$

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\begin{align} a-b & < a+b\\ \sqrt{a-b} & < \sqrt{a+b}\\ a-b & < \sqrt{a^2-b^2}\\ a- \sqrt{a^2-b^2} & < b\\ \frac{a- \sqrt{a^2-b^2}}{b}& < 1. \end{align}

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