| Statement-1:\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{2}^{1/x}}}{1+{{2}^{1/x}}}=1.\] |
| Statement-2:\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\cos }^{-1}}\left( 1-x \right)}{\sqrt{x}}=\sqrt{2}.\] |
A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
B) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
C) Statement-1 is true, Statement-2 is false.
D) Statement-1 is false, Statement-2 is true.
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{2}^{1/x}}}{1+{{2}^{1/x}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{1+{{2}^{-1/x}}}=1\] \[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{\cos }^{-1}}\left( 1-x \right)}{\sqrt{x}}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\theta }{\sqrt{1-\cos \theta }}\] (let,\[{{\cos }^{-1}}(1-x)=\theta \Rightarrow 1-x=\cos \theta )\]\[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{\theta }{\sqrt{2}\sin \left( \frac{\theta }{2} \right)}=\sqrt{2}\]
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