Limit

**Stroodle** · May 18th 2010, 02:40 AM

Hi there,

How would I evaluate

$\lim_{b\rightarrow\infty}\frac{a+b-\sqrt{a^2-ab+b^2}}{6}$

Thanks for your help.

**Archie Meade** · May 18th 2010, 03:44 AM

Originally Posted by Stroodle

Hi there,

How would I evaluate

$\lim_{b\rightarrow\infty}\frac{a+b-\sqrt{a^2-ab+b^2}}{6}$

Thanks for your help.

Hi Stroodle,

$\frac{a+b-\sqrt{a^2-ab+b^2}}{6}=\frac{a+b-\sqrt{\left(b-\frac{a}{2}\right)^2+\frac{3a^2}{4}}}{6}$

As $b\ \rightarrow\ \infty$ this approaches $\frac{a+b-\sqrt{\left(b-\frac{a}{2}\right)^2}}{6}$

$=\frac{a+b-b+\frac{a}{2}}{6}=\frac{3a}{12}$

**Stroodle** · May 18th 2010, 03:46 AM

Nice one. Thanks for that.

**galactus** · May 18th 2010, 03:53 AM

Rewrite it as:

$\frac{1}{6}\lim_{b\to {\infty}}\frac{a^{2}-ab}{-b-\sqrt{a^{2}-ab+b^{2}}}+\frac{a}{6}$

Now, divide the top by b, the b in the bottom by b and the radical by $\sqrt{b^{2}}$

$\frac{1}{6}\lim_{b\to {\infty}}\frac{\frac{a^{2}}{b}-\frac{ab}{b}}{\frac{-b}{b}-\sqrt{\frac{a^{2}}{b^{2}}-\frac{ab}{b^{2}}+\frac{b^{2}}{b^{2}}}}+\frac{a}{6}$

Then, it whittles down to:

$\frac{1}{6}\lim_{b\to {\infty}}\frac{-a}{-2}+\frac{a}{6}$

$\frac{a}{12}+\frac{a}{6}=\frac{a}{4}$

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