If lim _{x rightarrow infty}left(sqrt{x^{2}-x | vedclass

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(B) Given $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x\right)=b$. For the limit to exist as a finite value,the degree of the expression mus

Vedclass · Accepted Answer

$\left(1, -\frac{1}{2}\right)$

Vedclass · Answer

(B) Given $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x\right)=b$. For the limit to exist as a finite value,the degree of the expression must be balanced,so $a$ must be $1$. Rationalizing the expression: $\lim _{x \rightarrow \infty} \frac{(\sqrt{x^{2}-x+1}-ax)(\sqrt{x^{2}-x+1}+ax)}{\sqrt{x^{2}-x+1}+ax} = b$ $\lim _{x \rightarrow \infty} \frac{x^{2}-x+1-a^{2}x^{2}}{\sqrt{x^{2}-x+1}+ax} = b$ $\lim _{x \rightarrow \infty} \frac{(1-a^{2})x^{2}-x+1}{\sqrt{x^{2}-x+1}+ax} = b$ Since the limit is finite,the coefficient of $x^{2}$ must be $0$,so $1-a^{2}=0$. Since $x \rightarrow \infty$,we take $a=1$. Now,$\lim _{x \rightarrow \infty} \frac{-x+1}{\sqrt{x^{2}-x+1}+x} = b$ Dividing numerator and denominator by $x$: $\lim _{x}$ ${\rightarrow \infty} \frac{-1+\frac{1}{x}}{\sqrt{1-\frac{1}{x}+\frac{1}{x^{2}}}+1} = \frac{-1}{1+1} = -\frac{1}{2}$ Thus,$b = -\frac{1}{2}$. The ordered pair $(a, b)$ is $\left(1, -\frac{1}{2}\right)$.

If $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x\right)=b$,then the ordered pair $(a, b)$ is:

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