Evaluate the limit: lim _{x rightarrow 2} fra | vedclass

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(D) Given the limit: $\lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}$ Substituting $x = 2$,we get $\frac{2^3 - 2^2 - 2 - 2}{2(2^3) - 3(

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$\frac{7}{9}$

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(D) Given the limit: $\lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}$ Substituting $x = 2$,we get $\frac{2^3 - 2^2 - 2 - 2}{2(2^3) - 3(2^2) - 3(2) + 2} = \frac{8-4-2-2}{16-12-6+2} = \frac{0}{0}$ form. Applying $L$'$H$ôpital's rule by differentiating the numerator and denominator with respect to $x$: Numerator derivative: $\frac{d}{dx}(x^3-x^2-x-2) = 3x^2-2x-1$ Denominator derivative: $\frac{d}{dx}(2x^3-3x^2-3x+2) = 6x^2-6x-3$ Now,evaluate the limit: $\lim _{x \rightarrow 2} \frac{3x^2-2x-1}{6x^2-6x-3}$ $= \frac{3(2)^2 - 2(2) - 1}{6(2)^2 - 6(2) - 3} = \frac{12-4-1}{24-12-3} = \frac{7}{9}$

Evaluate the limit: $\lim _{x \rightarrow 2} \frac{x^3-x^2-x-2}{2 x^3-3 x^2-3 x+2}$

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