Question 975670
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Yep, you did it correctly:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\sqrt{x\ +\ h\ +\ 1}\ -\ sqrt{x\ +\ 1}}{h}]


When you rationalize the numerator, you end up with the difference of two squares in the numerator and a new factor in the denomiator that is the conjugate of the numerator.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{x\ +\ h\ +\ 1\ -\ (x\ +\ 1)}{h(\sqrt{x\ +\ h\ +\ 1}\ +\ sqrt{x\ +\ 1})}]


And finally:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{\sqrt{x\ +\ h\ +\ 1}\ +\ sqrt{x\ +\ 1}}]


which becomes


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


when you take the limit as *[tex \Large h\right 0]


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

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \