Functions Question 45
Question: True statement for $ \underset{x\to 0}{\mathop{\lim }},\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{2+3x}-\sqrt{2-3x}} $ is
[BIT Ranchi 1982]
Options:
A) Does not exist
B) Lies between 0 and $ \frac{1}{2} $
C) Lies between $ \frac{1}{2} $ and 1
D) Greater then 1
Show Answer
Answer:
Correct Answer: B
Solution:
$ \underset{x\to 0}{\mathop{\lim }}\frac{(1+x)-(1-x)}{(2+3x)-(2-3x)},[ \frac{\sqrt{2+3x}+\sqrt{2-3x}}{\sqrt{1+x}+\sqrt{1-x}} ] $ $ =\frac{1}{3},[ \frac{2\sqrt{2}}{2} ]=\frac{\sqrt{2}}{3},0<\frac{\sqrt{2}}{3}<\frac{1}{2}. $
Aliter : Apply L-Hospital?s rule,
$ \underset{x\to 0}{\mathop{\lim }},\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{2+3x}-\sqrt{2-3x}}=\underset{x\to 0}{\mathop{\lim }}\frac{\frac{1}{2\sqrt{1+x}}+\frac{1}{2\sqrt{1-x}}}{\frac{3}{2\sqrt{2+3x}}+\frac{3}{2\sqrt{2-3x}}} $
$ =\frac{\frac{1}{2}+\frac{1}{2}}{\frac{3}{2\sqrt{2}}+\frac{3}{2\sqrt{2}}}=\frac{2\sqrt{2}}{6}=\frac{\sqrt{2}}{3} $ .
BETA
