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 $\displaystyle \frac{a+b-\sqrt{\left(b-\frac{a}{2}\right)^2}}{6}$
$\displaystyle =\frac{a+b-b+\frac{a}{2}}{6}=\frac{3a}{12}$
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:
$\displaystyle \frac{1}{6}\lim_{b\to {\infty}}\frac{-a}{-2}+\frac{a}{6}$
$\displaystyle \frac{a}{12}+\frac{a}{6}=\frac{a}{4}$