A) 16
B) 8
C) 4
D) 2
Correct Answer: B
Solution :
[b] \[\underset{x\to 2}{\mathop{\lim }}\,\frac{{{2}^{x}}+{{2}^{3-x}}-6}{\sqrt{{{2}^{-x}}}-{{2}^{1-x}}}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{{{({{2}^{x}})}^{2}}-6\times {{2}^{x}}+{{2}^{3}}}{\sqrt{{{2}^{x}}}-2}-2\] [Multiplying \[{{N}^{r}}\]and \[{{D}^{r}}\]by\[{{2}^{x}}\]] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{({{2}^{x}}-4)({{2}^{x}}-2)(\sqrt{{{2}^{x}}}+2)}{(\sqrt{{{2}^{x}}}-2)(\sqrt{{{2}^{x}}}+2)}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{({{2}^{x}}-4)({{2}^{x}}-2)(\sqrt{{{2}^{x}}}+2)}{({{2}^{x}}-4)}\] \[=\underset{x\to 2}{\mathop{\lim }}\,({{2}^{x}}-2)(\sqrt{{{2}^{x}}}+2)=({{2}^{2}}-2)(2+2)=8\]
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