If lim _{x rightarrow infty}left(frac{x^2+1}{ | vedclass

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(B) Given $\lim _{x \rightarrow \infty} \left(\frac{x^2+1}{x+1}-ax-b\right) = 0$. Taking the common denominator: $\lim _{x \rightarrow \infty} \frac{x

Vedclass · Accepted Answer

$a=1, b=-1$

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(B) Given $\lim _{x \rightarrow \infty} \left(\frac{x^2+1}{x+1}-ax-b\right) = 0$. Taking the common denominator: $\lim _{x \rightarrow \infty} \frac{x^2+1-ax(x+1)-b(x+1)}{x+1} = 0$ $\lim _{x \rightarrow \infty} \frac{x^2+1-ax^2-ax-bx-b}{x+1} = 0$ $\lim _{x \rightarrow \infty} \frac{(1-a)x^2 - (a+b)x + (1-b)}{x+1} = 0$ For the limit to be $0$ as $x \rightarrow \infty$, the coefficient of the highest power of $x$ in the numerator must be $0$ and the degree of the numerator must be less than the degree of the denominator. Thus, $1-a = 0 \implies a = 1$. Substituting $a=1$ into the coefficient of $x$: $-(a+b) = 0 \implies -(1+b) = 0 \implies b = -1$. Therefore, $a=1$ and $b=-1$.

If $\lim _{x \rightarrow \infty}\left(\frac{x^2+1}{x+1}-a x-b\right)=0$, where $a, b \in R$, then:

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