lim _{x rightarrow 3} frac{x^3-27}{x^2-9} = | vedclass

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(B) To evaluate the limit $\lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9}$,we observe that substituting $x=3$ results in the indeterminate form $\frac{0

Vedclass · Accepted Answer

$\frac{9}{2}$

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(B) To evaluate the limit $\lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9}$,we observe that substituting $x=3$ results in the indeterminate form $\frac{0}{0}$.We factor the numerator using the difference of cubes formula $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and the denominator using the difference of squares formula $a^2-b^2 = (a-b)(a+b)$.$\lim _{x \rightarrow 3} \frac{(x-3)(x^2+3x+9)}{(x-3)(x+3)}$Canceling the common factor $(x-3)$,we get:$\lim _{x \rightarrow 3} \frac{x^2+3x+9}{x+3}$Substituting $x=3$:$\frac{3^2+3(3)+9}{3+3} = \frac{9+9+9}{6} = \frac{27}{6} = \frac{9}{2}$.

$\lim _{x \rightarrow 3} \frac{x^3-27}{x^2-9} = $

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