Algebra involved in limit...have the work, but need explanation

**janedoe** · September 11th 2010, 09:33 PM

http://i53.tinypic.com/vgr8yp.jpg

^see pic

Thank you so much!!

**Prove It** · September 11th 2010, 09:44 PM

They've multiplied by a cleverly disguised $1$ .

In this case, $\displaystyle\frac{\frac{1}{x}}{\frac{1}{x}}$ .

So $\displaystyle{\frac{x}{\sqrt{x^2 + x} + x} = \frac{x}{\sqrt{x^2+x}+x}\cdot \frac{\frac{1}{x}}{\frac{1}{x}}}$

$\displaystyle{ = \frac{1}{\frac{\sqrt{x^2+x}+x}{x}}}$

$\displaystyle{ = \frac{1}{\frac{\sqrt{x^2+x}}{x} + 1}}$

$\displaystyle{ = \frac{1}{\frac{\sqrt{x^2+x}}{\sqrt{x^2}} + 1}}$

$\displaystyle{= \frac{1}{\sqrt{\frac{x^2 + x}{x^2}} + 1}}$

$\displaystyle{= \frac{1}{\sqrt{1 + \frac{1}{x}} + 1}}$ .

**janedoe** · September 11th 2010, 10:04 PM

Ahh I see, thanks!! I wish I had been that perceptive and noticed that trick :[

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