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

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(A) The given function is a rational function. We first evaluate the function at $x = 2$. Substituting $x = 2$ gives $\frac{2^3 - 4(2^2) + 4(2)}{2^2 -

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(A) The given function is a rational function. We first evaluate the function at $x = 2$. Substituting $x = 2$ gives $\frac{2^3 - 4(2^2) + 4(2)}{2^2 - 4} = \frac{8 - 16 + 8}{4 - 4} = \frac{0}{0}$.Since it is an indeterminate form,we simplify the expression by factoring the numerator and the denominator.Numerator: $x^3 - 4x^2 + 4x = x(x^2 - 4x + 4) = x(x - 2)^2$.Denominator: $x^2 - 4 = (x - 2)(x + 2)$.Now,$\mathop {\lim }\limits_{x 	o 2} \frac{x(x-2)^2}{(x-2)(x+2)} = \mathop {\lim }\limits_{x 	o 2} \frac{x(x-2)}{x+2}$ (for $x 
eq 2$).Substituting $x = 2$ in the simplified expression: $\frac{2(2-2)}{2+2} = \frac{2(0)}{4} = \frac{0}{4} = 0$.

Find the limit: $\mathop {\lim }\limits_{x \to 2} \left[\frac{x^{3}-4 x^{2}+4 x}{x^{2}-4}\right]$

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