If mathop {lim }limits_{x to infty } left[ {f | vedclass

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Question

(C) Given $\mathop {\lim }\limits_{x 	o \infty } \left( {\frac{{{x^3} + 1}}{{{x^2} + 1}} - (ax + b)} ight) = 2$. Simplify the expression inside the

Vedclass · Accepted Answer

$a = 1$ and $b = -2$

Vedclass · Answer

(C) Given $\mathop {\lim }\limits_{x 	o \infty } \left( {\frac{{{x^3} + 1}}{{{x^2} + 1}} - (ax + b)} ight) = 2$. Simplify the expression inside the limit: $\frac{{{x^3} + 1 - (ax + b)({x^2} + 1)}}{{{x^2} + 1}} = \frac{{{x^3} + 1 - (a{x^3} + ax + b{x^2} + b)}}{{{x^2} + 1}} = \frac{{{x^3}(1 - a) - b{x^2} - ax + (1 - b)}}{{{x^2} + 1}}$. For the limit to be a finite value $(2)$,the coefficient of $x^3$ must be zero,so $1 - a = 0 \implies a = 1$. Now the expression becomes $\mathop {\lim }\limits_{x 	o \infty } \frac{{ - b{x^2} - x + (1 - b)}}{{{x^2} + 1}} = 2$. Divide numerator and denominator by $x^2$: $\mathop {\lim }\limits_{x 	o \infty } \frac{{ - b - \frac{1}{x} + \frac{{1 - b}}{{{x^2}}}}}{{1 + \frac{1}{{{x^2}}}}} = 2$. As $x 	o \infty$,this simplifies to $-b = 2$,so $b = -2$. Thus,$a = 1$ and $b = -2$.

If $\mathop {\lim }\limits_{x \to \infty } \left[ {\frac{{{x^3} + 1}}{{{x^2} + 1}} - (ax + b)} \right] = 2$,then

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