A) 3/2
B) 3
C) 2/3
D) 1/3
Correct Answer: B
Solution :
\[y=\underset{x\to 4}{\mathop{\lim }}\,\left[ \frac{{{x}^{3/2}}-8}{x-4} \right]\]\[=\underset{x\to 4}{\mathop{\lim }}\,\,\left[ \frac{{{({{x}^{1/2}})}^{3}}-{{(2)}^{3}}}{(\sqrt{x}-2)(\sqrt{x}+2)} \right]\] Þ \[y=\underset{x\to 4}{\mathop{\lim }}\,\frac{({{x}^{1/2}}-2)(x+4+2\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)}\] Þ \[y=\underset{x\to 4}{\mathop{\lim }}\,\frac{(x+4+2\sqrt{x})}{(\sqrt{x}+2)}\]\[=\frac{4+4+2\sqrt{4}}{\sqrt{4}+2}\]\[=\frac{12}{4}=3\]. Trick : Applying L-Hospital?s rule, we get \[\underset{x\to 4}{\mathop{\lim }}\,\frac{\frac{3}{2}{{x}^{1/2}}}{1}\]\[=\frac{3}{2}{{(4)}^{1/2}}\] = 3.
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