1.

7\lim _{x \rightarrow-3} \frac{2 x^{2}+7 x+3}{3 x^{2}+8 x-3}

Answer»

by putting x = -3 we will get 0/0 form so using L'hospital rule lim x->-3 (6x+8)/(4x+7) = (-18+8)/(-12+7) = -10/-5 = 2



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