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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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_

Vedclass · Accepted Answer

Has value $-1$

Vedclass · Answer

(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 $x$.

Similar Questions

$\mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^{ - 1}}x - {{\tan }^{ - 1}}x}}{{{x^3}}}$ is equal to

Which of the following limits vanish? (where $[ \cdot ]$ denotes the greatest integer function)

$\lim _{x \rightarrow 1} (\log _3 3x)^{\log _x 8} = \ldots$

$\lim _{x \rightarrow 2} \frac{x+3 x^2+5 x^3+7 x^4-166}{x-2} = $

$\mathop {\lim }\limits_{x \to \pi /2} \left[ {x\tan x - \left( {\frac{\pi }{2}} \right)\sec x} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module