Functions Question 178
Question: The value of $ \underset{x\to 2}{\mathop{\lim }},\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2} $ is
Options:
A) $ \underset{x\to \frac{\pi }{2}}{\mathop{\lim }},\frac{[ 1-\tan ( \frac{x}{2} ) ],[1-\sin x]}{[ 1+\tan ( \frac{x}{2} ) ],{{[\pi -2x]}^{3}}} $
B) $ \frac{1}{4\sqrt{3}} $
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We have $ \underset{x\to 2}{\mathop{\lim }}\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2} $ $ =\underset{x\to 2}{\mathop{\lim }},\frac{1+\sqrt{2+x}-3}{(\sqrt{1+\sqrt{2+x}+\sqrt{3})(x-2)}} $ $ =\underset{x\to 2}{\mathop{\lim }},\frac{\sqrt{2+x}-2}{(\sqrt{1+\sqrt{2+x}+\sqrt{3})(x-2)}} $ $ =\underset{x\to 2}{\mathop{\lim }},\frac{(x-2)}{(\sqrt{1+\sqrt{2+x}}+\sqrt{3})(\sqrt{2+x}+2)(x-2)} $ $ =\frac{1}{(2\sqrt{3}),4}=\frac{1}{8\sqrt{3}}. $
BETA
