Question

$ \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{8}}-2x+1}{{{x}^{4}}-2x+1} $ equals

• A. $ 3 $
• B. $ 0 $
• C. $ -3 $
• D. $ 1 $

Answer 1

Answer: (A)

$ \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 $