A) \[\frac{\cos \sqrt{x}}{2\sqrt{x}}\]
B) \[\sin \sqrt{x}\]
C) \[cos\sqrt{x}\]
D) \[\frac{1}{2}sin\sqrt{x}\]
Correct Answer: A
Solution :
[a] \[\because \underset{h\to 0}{\mathop{\lim }}\,\sin \frac{\sqrt{x+h}-\sin \sqrt{x}}{h}\] \[\underset{h\to 0}{\mathop{\lim }}\,\frac{2.\cos \frac{\sqrt{x+h}+\sqrt{x}}{2}.\sin \frac{\sqrt{x+h}-\sqrt{x}}{2}}{x+h-x}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2.\cos \frac{\sqrt{x+h}+\sqrt{x}}{2}.\sin \frac{\sqrt{x+h}-\sqrt{x}}{2}}{\left( \sqrt{x+h}+\sqrt{x} \right)\left( \sqrt{x+h}-\sqrt{x} \right)}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{2.\cos \sqrt{x+h}+\sqrt{x}}{2}\times \frac{\frac{\sqrt{x+h}-\sqrt{x}}{2}}{\frac{\sqrt{x+h}-\sqrt{x}}{2}}\times \frac{1}{2\left[ \sqrt{x+h}+\sqrt{x} \right]}\]\[=\underset{h\to 0}{\mathop{\lim }}\,2.\cos \frac{\left( \sqrt{x+h}+\sqrt{x} \right)}{2}\times \frac{1}{2\left[ \sqrt{x+h}+\sqrt{x} \right]}\] \[=2\left( \frac{\cos 2\sqrt{x}}{2} \right)\times \frac{1}{2(2\sqrt{x})}=\frac{\cos \sqrt{x}}{2\sqrt{x}}\]
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