Differentiation Question 491
Question: What is $ \underset{x\to 0}{\mathop{\lim }},\frac{x}{\sqrt{1-\cos x}} $ equal to?
Options:
A) $ \sqrt{2} $
B) $ -\sqrt{2} $
C) $ \frac{1}{\sqrt{2}} $
D) Limit does not exist
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ \underset{x\to 0}{\mathop{\lim }},\frac{x}{\sqrt{1-\cos x}} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{x}{\sqrt{1-( 1-2{{\sin }^{2}}\frac{x}{2} )}} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{x}{\sqrt{2{{\sin }^{2}}\frac{x}{2}}}=\frac{1}{2}\underset{x\to 0}{\mathop{\lim }},\frac{x}{| \sin \frac{x}{2} |} $ L.H.L $ =f(0-0)=\underset{h\to 0}{\mathop{\lim }},\frac{x}{| \sin \frac{x}{2} |} $ $ =-\frac{1}{\sqrt{2}}\underset{x\to 0}{\mathop{\lim }},\frac{2( \frac{h}{2} )}{\sin \frac{h}{2}} $ $ =\frac{1}{\sqrt{2}}\times 2\times 1 $ $ ( \because \underset{\theta \to 0}{\mathop{\lim }},\frac{\theta }{\sin \theta }=1 ) $ RHL $ =f(0+0)=\underset{h\to 0}{\mathop{\lim }},f(0+h) $ $ =\frac{1}{\sqrt{2}}\underset{h\to 0}{\mathop{\lim }},\frac{2( \frac{h}{2} )}{\sin \frac{h}{2}}=\frac{1}{\sqrt{2}}\times 2\times 1 $ $ LHL\ne RHL=\sqrt{2} $ Therefore limit does not exist.
BETA
