A) \[3\]
B) \[0\]
C) \[-3\]
D) \[1\]
Correct Answer: A
Solution :
\[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{8}}-2x+1}{{{x}^{4}}-2x+1}\left( \frac{0}{0}\text{form} \right)\] \[\Rightarrow \]\[\underset{x\to 1}{\mathop{\lim }}\,\frac{8{{x}^{7}}-2}{4{{x}^{3}}-2}=\frac{8-2}{4-2}=3\] (using L' Hospital's rule) Alternate Method: Using factorisation method \[\therefore \] \[\underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{8}}-2x+1}{{{x}^{4}}-2x+1}\] \[\underset{x\to 1}{\mathop{\lim }}\,\frac{({{x}^{7}}+{{x}^{6}}+{{x}^{5}}+{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x-1)(x-1)}{({{x}^{3}}+{{x}^{2}}+x-1)(x-1)}\] \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{({{x}^{7}}+{{x}^{6}}+{{x}^{5}}+{{x}^{4}}+{{x}^{3}}+{{x}^{2}}+x-1)}{{{x}^{3}}+{{x}^{2}}+x-1}\] \[=\frac{7-1}{3-1}=\frac{6}{2}=3\]
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