Functions Question 27
Question: $ \underset{\alpha \to \pi /4}{\mathop{\lim }}\frac{\sin \alpha -\cos \alpha }{\alpha -\frac{\pi }{4}}= $
[IIT 1977]
Options:
A) $ \sqrt{2} $
B) $ 1/\sqrt{2} $
C) 1
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{\alpha \to \pi /4}{\mathop{\lim }}\frac{\sin \alpha -\cos \alpha }{\alpha -\pi /4} $ $ =\underset{\alpha \to \pi /4}{\mathop{\lim }}{ \frac{\sqrt{2}( \sin \alpha .\frac{1}{\sqrt{2}}-\cos \alpha .\frac{1}{\sqrt{2}} )}{( \alpha -\frac{\pi }{4} )} } $ $ =\sqrt{2}\underset{\alpha \to \pi /4}{\mathop{\lim }}\frac{\sin ( \alpha -\frac{\pi }{4} )}{( \alpha -\frac{\pi }{4} )}=\sqrt{2}\times 1=\sqrt{2} $ . Aliter : Apply L-Hospital?s rule, $ \underset{\alpha \to \pi /4}{\mathop{\lim }}\frac{\sin \alpha -\cos \alpha }{\alpha -(\pi /4)}=\underset{\alpha \to \pi /4}{\mathop{\lim }}\frac{\cos \alpha +\sin \alpha }{1}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} $ .
BETA
