The value of mathop {lim }limits_{x to infty | vedclass

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(B) Let $x = \frac{1}{h}$. As $x 	o \infty$,$h 	o 0^+$.The expression becomes $\mathop {\lim }\limits_{h 	o 0} \left( \frac{1}{h^2} + \frac{1}{h} \

Vedclass · Accepted Answer

$-\frac{1}{3}$

Vedclass · Answer

(B) Let $x = \frac{1}{h}$. As $x 	o \infty$,$h 	o 0^+$.The expression becomes $\mathop {\lim }\limits_{h 	o 0} \left( \frac{1}{h^2} + \frac{1}{h} ight) \log \left( \frac{1}{h} \cot^{-1} \frac{1}{h} ight) = \mathop {\lim }\limits_{h 	o 0} \left( \frac{1+h}{h^2} ight) \log \left( \frac{	an^{-1} h}{h} ight)$.Using the expansion $	an^{-1} h = h - \frac{h^3}{3} + \frac{h^5}{5} - \dots$,we have $\frac{	an^{-1} h}{h} = 1 - \frac{h^2}{3} + \frac{h^4}{5} - \dots$.Thus,the limit is $\mathop {\lim }\limits_{h 	o 0} \left( \frac{1+h}{h^2} ight) \log \left( 1 - \frac{h^2}{3} + \frac{h^4}{5} - \dots ight)$.Using $\log(1+u) \approx u$ for small $u$,we get $\mathop {\lim }\limits_{h 	o 0} \left( \frac{1+h}{h^2} ight) \left( -\frac{h^2}{3} + \frac{h^4}{5} ight) = \mathop {\lim }\limits_{h 	o 0} (1+h) \left( -\frac{1}{3} + \frac{h^2}{5} ight) = -\frac{1}{3}$.

The value of $\mathop {\lim }\limits_{x \to \infty } \left( {\left| {{x^2}} \right| + x} \right)\log \left( {x{{\cot }^{ - 1}}x} \right)$ is

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