Finding Limits

**frenchguy87** · February 3rd 2010, 04:37 PM

I need to find the lim (squ(1+2x)-squ(1+3x)) / (x+2x^2) as x goes to 0 and where x>0.

Would you multiply by the conjugate?

**xalk** · February 3rd 2010, 05:10 PM

Originally Posted by frenchguy87

I need to find the lim (squ(1+2x)-squ(1+3x)) / (x+2x^2) as x goes to 0 and where x>0.

Would you multiply by the conjugate?

Do you mean the limit of:

$\frac{(1+2x)^2-(1+3x)^2}{x+2x^2}$ as x goes to zero??

**mr fantastic** · February 3rd 2010, 05:23 PM

Originally Posted by frenchguy87

I need to find the lim (squ(1+2x)-squ(1+3x)) / (x+2x^2) as x goes to 0 and where x>0.

Would you multiply by the conjugate?

$\lim_{x \to 0} \frac{\sqrt{1 + 2x} - \sqrt{1 + 3x}}{x + 2x^2} = \lim_{x \to 0} \frac{(\sqrt{1 + 2x} - \sqrt{1 + 3x})}{(x + 2x^2)} \cdot \frac{ (\sqrt{1 + 2x} + \sqrt{1 + 3x}) }{(\sqrt{1 + 2x} + \sqrt{1 + 3x}) }$

$= \lim_{x \to 0} \frac{-x}{(x + 2x^2) (\sqrt{1 + 2x} + \sqrt{1 + 3x}) } = ....$

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