# Limits of indeterminant form.

• January 9th 2011, 02:38 PM
niazk90
Limits of indeterminant form.
• January 9th 2011, 11:08 PM
mr fantastic
Quote:

Originally Posted by niazk90

For Q3 note that $\displaystyle \frac{\sqrt{x^2 - x}}{x-9} = \frac{\sqrt{1 - \frac{1}{x}}}{1 - \frac{9}{x}}$.

For Q6 note that by multiplying by $\displaystyle \frac{\sqrt{2x+3} + \sqrt{x - 1}}{\sqrt{2x+3} + \sqrt{x - 1}}$ the expression becomes $\displaystyle \frac{x + 4}{\sqrt{2x+3} + \sqrt{x - 1}}$.