Thread: Limit

Hi there,

How would I evaluate

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

Thanks for your help.

Originally Posted by Stroodle

Hi there,

How would I evaluate

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

Thanks for your help.

Hi Stroodle,

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

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

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

Nice one. Thanks for that.

Rewrite it as:

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

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

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

Then, it whittles down to:

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

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