A) \[\frac{{{\log }_{2}}-1}{{{\log }_{2}}+1}\]
B) \[\pm 1\]
C) \[\frac{{{\log }_{2}}+1}{{{\log }_{2}}-1}\]
D) \[-1\]
Correct Answer: A
Solution :
[a] \[\therefore \underset{x\to 2}{\mathop{\lim }}\,\frac{{{2}^{x}}-{{x}^{2}}}{{{x}^{x}}-{{2}^{2}}}\] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{{{e}^{x\log 2}}-{{x}^{2}}}{{{e}^{x\log x}}-4}\] which is \[\frac{0}{0}\]from \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{\log 2.{{e}^{x\log 2}}-2x}{{{e}^{x\log x}}\left( x.\frac{1}{x}+\log x \right)-0}\] [By LH. Rule] \[=\underset{x\to 2}{\mathop{\lim }}\,\frac{\log 2.{{(e)}^{x\log 2}}-2x}{{{e}^{x\log x}}(1+\log x)}\] Applying limit, \[x\to 2\], we have \[=\frac{\log {{2.2}^{2}}-2\times 2}{{{2}^{2}}(1+\log 2)}=\frac{4(\log 2-1)}{4(1+\log 2)}=\frac{\log 2-1}{\log 2+1}\] Hence, option [a] is correct.
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