A) 1
B) \[\frac{1}{1-x}\]
C) \[\frac{1}{1+x}\]
D) \[\frac{-1}{{{(1-x)}^{2}}}\]
E) \[\frac{1}{{{(1-x)}^{2}}}\]
Correct Answer: E
Solution :
\[y=\underset{n\to \infty }{\mathop{\lim }}\,(1+x){{(1+x)}^{2}}(1+{{x}^{4}})....(1+{{x}^{{{2}^{n}}}})\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{(1-x)(1+x)(1+{{x}^{2}})(1+{{x}^{4}})....(1+{{x}^{{{2}^{n}}}})}{(1-x)}\] \[=\underset{n\to \infty }{\mathop{\lim }}\,\frac{(1-{{x}^{2}})(1+{{x}^{2}})(1+{{x}^{4}})....(1+{{x}^{{{2}^{n}}}})}{1-x}\] \[\Rightarrow \] \[y=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1-{{x}^{{{2}^{n+1}}}}}{1-x}=\frac{1}{{{(1-x)}^{2}}}\] \[\Rightarrow \] \[y=\frac{1}{{{(1-x)}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
