lim _{x rightarrow 3^{-}} frac{x^3-3 x^2-4 x+ | vedclass

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(C) Given,$L = \lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3}$.Checking the form at $x = 3$,we get $\frac{3^3 - 3(3^2) - 4(3)

Vedclass · Accepted Answer

$\frac{5}{14}$

Vedclass · Answer

(C) Given,$L = \lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3}$.Checking the form at $x = 3$,we get $\frac{3^3 - 3(3^2) - 4(3) + 12}{2(3^3) - 7(3^2) + 2(3) + 3} = \frac{27 - 27 - 12 + 12}{54 - 63 + 6 + 3} = \frac{0}{0}$ form.Applying $L'H\hat{o}pital's$ rule,we differentiate the numerator and denominator with respect to $x$:$L = \lim _{x \rightarrow 3^{-}} \frac{\frac{d}{dx}(x^3-3 x^2-4 x+12)}{\frac{d}{dx}(2 x^3-7 x^2+2 x+3)} = \lim _{x \rightarrow 3^{-}} \frac{3 x^2-6 x-4}{6 x^2-14 x+2}$.Now,substitute $x = 3$:$L = \frac{3(3^2) - 6(3) - 4}{6(3^2) - 14(3) + 2} = \frac{27 - 18 - 4}{54 - 42 + 2} = \frac{5}{14}$.

$\lim _{x \rightarrow 3^{-}} \frac{x^3-3 x^2-4 x+12}{2 x^3-7 x^2+2 x+3} = $

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