A) \[\frac{1}{\sqrt{2}}\]
B) \[\frac{1}{2\sqrt{2}}\]
C) \[\frac{\sqrt{3}}{2}\]
D) \[\sqrt{3}\]
Correct Answer: A
Solution :
\[\underset{x\to 3}{\mathop{\lim }}\,\frac{\sqrt{3x}-3}{\sqrt{2x-4}-\sqrt{2}}\] Rationalize \[\underset{x\to 3}{\mathop{\lim }}\,\frac{(3x-9)\times \left( \sqrt{2x-4}+\sqrt{2} \right)}{\{\left( 2x-4 \right)-2\}\times \left( \sqrt{3x}+3 \right)}\] \[=\underset{x\to 3}{\mathop{\lim }}\,\frac{\left( 3x-3 \right)}{2\left( x-3 \right)}\times \frac{\sqrt{2x-4}+\sqrt{2}}{\left( \sqrt{3}x+3 \right)}\] \[=\frac{3}{2}\times \frac{2\sqrt{2}}{6}=\frac{1}{\sqrt{2}}\]
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