Evaluate mathop {lim }limits_{x to infty } fr | vedclass

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(A) We are given the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{\log {x^n} - [x]}}{{[x]}}.$This can be rewritten as $\mathop {\lim }\limits_

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Has value $-1$

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(A) We are given the limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{\log {x^n} - [x]}}{{[x]}}.$This can be rewritten as $\mathop {\lim }\limits_{x 	o \infty } \left( \frac{{\log {x^n}}}{{[x]}} - \frac{{[x]}}{{[x]}} ight).$Since $x - 1 Thus,$\mathop {\lim }\limits_{x 	o \infty } \frac{{\log {x^n}}}{{[x]}} = \mathop {\lim }\limits_{x 	o \infty } \frac{{n \log x}}{x}.$Using $L$'Hopital's rule or the standard limit $\mathop {\lim }\limits_{x 	o \infty } \frac{{\log x}}{x} = 0,$ we get $n 	imes 0 = 0.$Therefore,the expression becomes $0 - 1 = -1.$

Evaluate $\mathop {\lim }\limits_{x \to \infty } \frac{{\log {x^n} - [x]}}{{[x]}},$ where $n \in N$ and $[x]$ denotes the greatest integer less than or equal to $x$.

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