Differentiation Question 31
Question: The value of $ \underset{x\to 2}{\mathop{\lim }}\frac{2^{x}+{2^{3-x}}-6}{\sqrt{{2^{-x}}}-{2^{1-x}}} $ is
Options:
A) 16
B) 8
C) 4
D) 2
Show Answer
Answer:
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 $
BETA
