• # question_answer $\underset{h\to 0}{\mathop{Lt}}\,\frac{\mathbf{sin}\sqrt{\mathbf{x}+\mathbf{h}}-\mathbf{sin}\sqrt{\mathbf{x}}}{h}$ is equal to A) $\frac{\cos \sqrt{x}}{2\sqrt{x}}$                                   B) $\sin \sqrt{x}$         C) $cos\sqrt{x}$                           D)  $\frac{1}{2}sin\sqrt{x}$

[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]}$ $=2\left( \frac{\cos 2\sqrt{x}}{2} \right)\times \frac{1}{2(2\sqrt{x})}=\frac{\cos \sqrt{x}}{2\sqrt{x}}$