Find the limit: mathop {lim }limits_{x to 2} | vedclass

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Question

(A) The given function is a rational function. We first evaluate the function at $x = 2$. Substituting $x = 2$ in the expression,we get $\frac{2^3 - 2

Vedclass · Accepted Answer

$-4$

Vedclass · Answer

(A) The given function is a rational function. We first evaluate the function at $x = 2$. Substituting $x = 2$ in the expression,we get $\frac{2^3 - 2(2)^2}{2^2 - 5(2) + 6} = \frac{8 - 8}{4 - 10 + 6} = \frac{0}{0}$.Since the form is $\frac{0}{0}$,we simplify the expression by factoring the numerator and the denominator.$\mathop {\lim }\limits_{x 	o 2} \frac{x^2(x-2)}{(x-2)(x-3)}$Canceling the common factor $(x-2)$ for $x 
eq 2$,we get:$\mathop {\lim }\limits_{x 	o 2} \frac{x^2}{x-3}$Now,substituting $x = 2$:$\frac{2^2}{2-3} = \frac{4}{-1} = -4$.

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{3}-2 x^{2}}{x^{2}-5 x+6}\right]$

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