A) \[f(x)=\frac{1}{In\left| x \right|}\]
B) \[f(x)=\frac{1}{{{x}^{3}}-1}\]
C) \[f(x)={{2}^{{{2}^{\frac{1}{1-x}}}}}\]
D) \[f(x)=\frac{\sqrt{x+1}-\sqrt{2x}}{{{x}^{2}}-x}\]
Correct Answer: D
Solution :
[d] [a] \[\underset{x\to 1}{\mathop{\lim }}\,f(x)\] does not exist. [b] \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=\] does not exist. [c] \[\underset{x\to 1}{\mathop{\lim }}\,f(x)\] does not exist. [d] \[\underset{x\to 1}{\mathop{\lim }}\,f(x)=\frac{-1}{2\sqrt{2}},\] therefore \[f(x)\] has removable discontinuity at \[x=1\].
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