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Determine the following limits the limit as x approaches 9 of (9-x)/(3-sqrtx)
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\[\large \lim_{x\rightarrow 9} \frac{9-x}{3-\sqrt{x}}\] Multiply the numerator and denominator by the conjugate.
So...\[\lim_{x\rightarrow 9} \left[\frac{9-x}{3-\sqrt{x}} \cdot \frac{3+\sqrt{x}}{3+\sqrt{x}}\right]\]
rewrite the numerator as the difference of 2 squares \[\lim_{x \rightarrow 9} \frac{9 - (\sqrt{x})^2}{3 - \sqrt{x}}\] factor the numerator... and things will become obvious...
Ahh...several methods of approach!
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