A) \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\frac{\left[ 1-\tan \left( \frac{x}{2} \right) \right]\,[1-\sin x]}{\left[ 1+\tan \left( \frac{x}{2} \right) \right]\,{{[\pi -2x]}^{3}}}\]
B) \[\frac{1}{4\sqrt{3}}\]
C) 0
D) None of these
Correct Answer: A
Solution :
We have \[\underset{x\to 2}{\mathop{\lim }}\,\,\frac{\sqrt{1+\sqrt{2+x}}-\sqrt{3}}{x-2}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\,\,\frac{1+\sqrt{2+x}-3}{(\sqrt{1+\sqrt{2+x}+\sqrt{3})\,\,(x-2)}}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\,\,\frac{\sqrt{2+x}-2}{(\sqrt{1+\sqrt{2+x}+\sqrt{3})\,\,(x-2)}}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\,\,\frac{(x-2)}{(\sqrt{1+\sqrt{2+x}}+\sqrt{3})\,\,(\sqrt{2+x}+2)\,\,(x-2)}\] \[=\frac{1}{(2\sqrt{3})\,4}=\frac{1}{8\sqrt{3}}.\]
You need to login to perform this action.
You will be redirected in
3 sec
