# Limits of indeterminant form.

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• January 9th 2011, 02:38 PM
niazk90
Limits of indeterminant form.
Attachment 20372Please help with questions 3 and 6.
• January 9th 2011, 11:08 PM
mr fantastic
Quote:

Originally Posted by niazk90
Attachment 20372Please help with questions 3 and 6.

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 $\displaystyle \frac{x + 4}{\sqrt{2x+3} + \sqrt{x - 1}}$.