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