If lim _{x rightarrow 2} frac{3 x^2-a x+5 b}{ | vedclass

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(D) Given $\lim _{x \rightarrow 2} \frac{3 x^2-a x+5 b}{x-2}=17$. Since the limit exists and the denominator approaches $0$ as $x \rightarrow 2$,the n

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$22$

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(D) Given $\lim _{x ightarrow 2} \frac{3 x^2-a x+5 b}{x-2}=17$. Since the limit exists and the denominator approaches $0$ as $x ightarrow 2$,the numerator must also approach $0$ at $x=2$. $3(2)^2 - a(2) + 5b = 0$ $\Rightarrow 12 - 2a + 5b = 0$ $\Rightarrow 2a - 5b = 12$. Applying $L$'Hospital's rule: $\lim _{x ightarrow 2} \frac{d}{dx}(3x^2 - ax + 5b) / \frac{d}{dx}(x-2) = 17$. $\lim _{x}$ ${ightarrow 2} (6x - a) = 17$ $\Rightarrow 6(2) - a = 17$ $\Rightarrow 12 - a = 17$ $\Rightarrow a = -5$. Substitute $a = -5$ into $2a - 5b = 12$: $2(-5) - 5b = 12$ $\Rightarrow -10 - 5b = 12$ $\Rightarrow -5b = 22$ $\Rightarrow b = -\frac{22}{5}$. Therefore,$ab = (-5) 	imes (-\frac{22}{5}) = 22$.

If $\lim _{x \rightarrow 2} \frac{3 x^2-a x+5 b}{x-2}=17$,then $a b=$

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