Limit

Jun 2009
367
122
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.
 
Dec 2009
3,120
1,342
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 \(\displaystyle \frac{a+b-\sqrt{\left(b-\frac{a}{2}\right)^2}}{6}\)

\(\displaystyle =\frac{a+b-b+\frac{a}{2}}{6}=\frac{3a}{12}\)
 
  • Like
Reactions: Stroodle

galactus

MHF Hall of Honor
Jul 2006
3,002
1,124
Chaneysville, PA
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:

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

\(\displaystyle \frac{a}{12}+\frac{a}{6}=\frac{a}{4}\)
 
  • Like
Reactions: Stroodle