If lim _{x rightarrow 1} frac{x^2-ax+b}{x-1}= | vedclass

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(D) Given $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=5$. Since the limit exists and the denominator approaches $0$ as $x \rightarrow 1$,the numerat

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$-7$

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(D) Given $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=5$. Since the limit exists and the denominator approaches $0$ as $x \rightarrow 1$,the numerator must also approach $0$ for the limit to be finite. Thus,$1^2 - a(1) + b = 0$ $\Rightarrow 1 - a + b = 0$ $\Rightarrow b = a - 1$. Substituting $b$ into the limit: $\lim _{x \rightarrow 1} \frac{x^2 - ax + (a-1)}{x-1} = 5$. Using $L$'Hopital's rule: $\lim _{x \rightarrow 1} \frac{2x - a}{1} = 5$. $2(1) - a = 5$ $\Rightarrow 2 - a = 5$ $\Rightarrow a = -3$. Since $b = a - 1$,$b = -3 - 1 = -4$. Therefore,$a + b = -3 + (-4) = -7$.

If $\lim _{x \rightarrow 1} \frac{x^2-ax+b}{x-1}=5$,then $(a+b)$ is equal to

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