If lim _{x rightarrow 4} frac{2 x^2+(3+2 a) x | vedclass

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(D) Given $\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60} = \frac{11}{9}$.Since the limit exists,the numerator must be $0$ at $

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$-\frac{1}{9}$

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(D) Given $\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60} = \frac{11}{9}$.Since the limit exists,the numerator must be $0$ at $x=4$: $2(16) + (3+2a)(4) + 3a = 0$.$32 + 12 + 8a + 3a = 0 \Rightarrow 11a = -44 \Rightarrow a = -4$.Alternatively,using $L$'Hospital's Rule: $\lim _{x \rightarrow 4} \frac{4x + 3 + 2a}{3x^2 - 4x - 23} = \frac{16 + 3 + 2a}{48 - 16 - 23} = \frac{19 + 2a}{9} = \frac{11}{9}$.$19 + 2a = 11 \Rightarrow 2a = -8 \Rightarrow a = -4$.Now,evaluate $\lim _{x \rightarrow -4} \frac{x^2+9x+20}{x^2-x-20} = \lim _{x \rightarrow -4} \frac{(x+4)(x+5)}{(x+4)(x-5)} = \lim _{x \rightarrow -4} \frac{x+5}{x-5}$.Substituting $x = -4$: $\frac{-4+5}{-4-5} = \frac{1}{-9} = -\frac{1}{9}$.

If $\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}=\frac{11}{9}$,then $\lim _{x \rightarrow a} \frac{x^2+9 x+20}{x^2-x-20}=$

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