A) \[\frac{3}{8}\]
B) \[\frac{3}{2}\]
C) \[\frac{4}{3}\]
D) \[\frac{8}{3}\]
Correct Answer: D
Solution :
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{4}}-1}{x-1}=\underset{x\to k}{\mathop{\lim }}\,\frac{{{x}^{3}}-{{k}^{3}}}{{{x}^{2}}-{{k}^{2}}},\] \[\Rightarrow \underset{x\to 1}{\mathop{\lim }}\,\left( x+1 \right)\left( {{x}^{2}}+1 \right)=\frac{{{k}^{2}}+{{k}^{2}}+{{k}^{2}}}{2k}\] \[\Rightarrow k=8/3\]
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