Question 642347
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The trick here is to deny our instincts and rationalize the <i><b>numerator</b></i>


Note that the numerator is the difference of two quantities that we would very much like to square individually.  Recalling that the product of a pair of conjugates is the difference of two squares, multiply the fraction by 1 in the form of the conjugate of the numerator divided by itself:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \lim_{x\rightarrow{h}}\,\left( \frac{\sqrt{x\ +\ h}\ -\ \sqrt{x} }{h}\right)\left(\frac{\sqrt{x\ +\ h}\ +\ \sqrt{x}}{\sqrt{x\ +\ h}\ +\ \sqrt{x}}\right)\ =\ \lim_{x\rightarrow{h}}\,\frac{x\ +\ h\ -\ x}{h\left(\sqrt{x\ +\ h}\ +\ \sqrt{x})}\ =\ \lim_{x\rightarrow{h}}\,\frac{h}{h\left(\sqrt{x\ +\ h}\ +\ \sqrt{x}\right)}\ =\ \lim_{x\rightarrow{h}}\,\frac{1}{\left(\sqrt{x\ +\ h}\ +\ \sqrt{x}\right)]


Now all you have to do is drive *[tex \LARGE h] to zero:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ =\ \frac{1}{2\sqrt{x}}]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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