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

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Question

(A) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } (\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})$,we rationalize the expressio

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } (\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})$,we rationalize the expression:$= \mathop {\lim }\limits_{x 	o \infty } \frac{(\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})(\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1})}{\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1}}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{({a^2}{x^2} + ax + 1) - ({a^2}{x^2} + 1)}{\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1}}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{ax}{\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1}}$Dividing the numerator and denominator by $x$ (assuming $x > 0$ as $x 	o \infty$):$= \mathop {\lim }\limits_{x 	o \infty } \frac{a}{\sqrt {{a^2} + \frac{a}{x} + \frac{1}{{{x^2}}}} + \sqrt {{a^2} + \frac{1}{{{x^2}}}}}$As $x 	o \infty$,$\frac{1}{x} 	o 0$ and $\frac{1}{{{x^2}}} 	o 0$:$= \frac{a}{\sqrt {{a^2}} + \sqrt {{a^2}}} = \frac{a}{2|a|}$If $a > 0$,the limit is $\frac{a}{2a} = \frac{1}{2}$. If $a  0$ based on the options,the answer is $\frac{1}{2}$.

The value of $\mathop {\lim }\limits_{x \to \infty } (\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})$ is

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