Functions Question 88
Question: $ \underset{x\to a}{\mathop{\lim }},\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}= $
[IIT 1978; Kurukshetra CEE 1996]
Options:
A) $ \frac{1}{\sqrt{3}} $
B) $ \frac{2}{3\sqrt{3}} $
C) $ \frac{2}{\sqrt{3}} $
D) $ \frac{2}{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{x\to a}{\mathop{\lim }}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}} $ $ =\underset{x\to a}{\mathop{\lim }}\frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}\times \frac{\sqrt{a+2x}+\sqrt{3x}}{\sqrt{a+2x}+\sqrt{3x}}\times \frac{\sqrt{3a+x}+2\sqrt{x}}{\sqrt{3a+x}+2\sqrt{x}} $ $ =\underset{x\to a}{\mathop{\lim }},\frac{\sqrt{3a+x}+2\sqrt{x}}{3,(\sqrt{a+2x}+\sqrt{3x)}}=\frac{2}{3\sqrt{3}} $ . Aliter : Apply L-Hospital?s rule.
BETA
