A) 0
B) \[n{{a}^{n-1}}\]
C) \[n{{a}^{n}}\]
D) \[\infty \]
Correct Answer: B
Solution :
\[\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{n}}-{{a}^{n}}}{x-a}=\underset{x\to a}{\mathop{\lim }}\,\frac{n{{x}^{n-1}}-0}{1}=n.{{a}^{n-1}}\]You need to login to perform this action.
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