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Question: $$\frac{\frac{1}{\sqrt {x+h}}- \frac{1}{\sqrt x}}{h}$$

Solution given:

$$= \frac{1}{h} \cdot\frac{\sqrt x - \sqrt {x+h}} {\sqrt {x + h}\sqrt x} $$

$$= \frac{1}{h} \cdot\ \frac{x - (x+h)}{\sqrt{x + h} \sqrt x (\sqrt{x} + \sqrt{x + h})}$$

$$= \frac{1}{h} \cdot\ \frac{x - x - h}{x \sqrt{x + h} + (x + h) \sqrt x}$$

$$= \frac{1}{h} \cdot\ \frac{-h}{x \sqrt{x + h} + (x + h) \sqrt x}$$

$$= -\frac{1}{x \sqrt{x + h} + (x + h) \sqrt x}$$

I've studied and understood the material up to this point just fine. I get about rationalizing stuff, conjugate pairs etc, but I can't figure out of what the author has between each step to get to the next.

I can only comprehend the first and possibly the second step. Source: http://www.themathpage.com/alg/multiply-radicals.htm

See problem 10, the last problem on the page.

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    $\begingroup$ What step is in question? If you understand "rationalizing stuff," then this development is straightforward. By the way, the last step requires a minus sign. $\endgroup$ – Mark Viola Sep 25 '16 at 0:09
  • $\begingroup$ I guess... I can rationalize the denominators of simpler fractional radicals, but this one is confusing to me second step onward. I think in the second step he is rationalizing the denominator, but it looks odd. I know I'm not articulating myself well... $\endgroup$ – Shiny_and_Chrome Sep 25 '16 at 0:14
  • $\begingroup$ I don't really see how the last line is more simplified than the original expression, personally. $\endgroup$ – GFauxPas Sep 25 '16 at 0:16
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    $\begingroup$ @GFauxPas It permits ease of evaluating the limit as $-\frac{1}{2\sqrt{x}}$, whereas the original expression is of indeterminate form. And use of LHR would be circular logic. $\endgroup$ – Mark Viola Sep 25 '16 at 0:19
  • $\begingroup$ @Dr.MV ah, nice. $\endgroup$ – GFauxPas Sep 25 '16 at 0:31
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Step 1: Delete the chain break and multiply with the denominator of the other fraciton

Step 2: Use Binomial formula since $(a-b)(a+b)=a^2-b^2$

Step 3: Use Associative law of addition multiply everything out in the denominator

Step 4: $(x-x)=0$

Step 5: $\frac{-h}{h} = -1$

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  • $\begingroup$ Fantastic. I followed your instructions exactly it all made sense. Thank you. Just takes a while to recognise the form when the expressions start to get bigger. $\endgroup$ – Shiny_and_Chrome Sep 25 '16 at 0:35

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