lim _{x rightarrow infty}left[sqrt{x^2+a x+b} | vedclass

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(C) To evaluate the limit $\lim _{x \rightarrow \infty}\left(\sqrt{x^2+a x+b}-x\right)$,we rationalize the expression:$= \lim _{x}$ ${\rightarrow \inf

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depends only on $a$

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(C) To evaluate the limit $\lim _{x ightarrow \infty}\left(\sqrt{x^2+a x+b}-xight)$,we rationalize the expression:$= \lim _{x}$ ${ightarrow \infty}\left(\left(\sqrt{x^2+a x+b}-xight) 	imes \frac{\sqrt{x^2+a x+b}+x}{\sqrt{x^2+a x+b}+x}ight)$$= \lim _{x ightarrow \infty}\left(\frac{x^2+a x+b-x^2}{\sqrt{x^2+a x+b}+x}ight)$$= \lim _{x ightarrow \infty}\left(\frac{a x+b}{\sqrt{x^2+a x+b}+x}ight)$Divide the numerator and denominator by $x$:$= \lim _{x ightarrow \infty} \frac{a+\frac{b}{x}}{\sqrt{1+\frac{a}{x}+\frac{b}{x^2}}+1}$As $x ightarrow \infty$,$\frac{1}{x} ightarrow 0$ and $\frac{1}{x^2} ightarrow 0$,so the limit becomes:$= \frac{a+0}{\sqrt{1+0+0}+1} = \frac{a}{2}$Since the result is $\frac{a}{2}$,the limit depends only on $a$.

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+a x+b}-x\right]$ where $a < 0 < b$.

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