A) 0
B) 1
C) ¥
D) \[-1\]
Correct Answer: B
Solution :
[b] \[\because \underset{x\to 1}{\mathop{\lim }}\,\frac{{{x}^{7}}-2{{x}^{5}}+1}{{{x}^{3}}-3{{x}^{2}}+2}\] which is \[\frac{0}{0}\]form \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{\frac{d}{dx}\left( {{x}^{7}}-2{{x}^{5}}+1 \right)}{\frac{d}{dx}\left( {{x}^{3}}-3{{x}^{2}}+2 \right)}\] [By L. Hospital Rule] \[=\underset{x\to 1}{\mathop{\lim }}\,\frac{7{{x}^{6}}-10{{x}^{4}}}{3{{x}^{2}}-6x}=\frac{7-10}{3-6}=\frac{-3}{-3}=1\] Hence, option [b]is correct.You need to login to perform this action.
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