evaluate the limit. if the limit does not exist enter dne. limit = lim(x→3) (4x² - 4x + 8)/(x - 3)

Explanation:

Step1: Substitute $x = 3$ into the function.

When we substitute $x = 3$ into $\frac{4x^{2}-4x + 8}{x - 3}$, we get $\frac{4\times3^{2}-4\times3 + 8}{3 - 3}=\frac{4\times9-12 + 8}{0}=\frac{36-12 + 8}{0}=\frac{32}{0}$, which is undefined. So, we need to simplify the function if possible. Since the numerator $4x^{2}-4x + 8$ cannot be factored in a way that cancels out the denominator $x - 3$, we consider the one - sided limits.

Step2: Consider the limit as $x\to3^{+}$ and $x\to3^{-}$.

As $x\to3^{+}$, the numerator $4x^{2}-4x + 8=4x(x - 1)+8>0$ (when $x$ is close to 3) and the denominator $x - 3>0$ and approaches 0. So, $\lim_{x\to3^{+}}\frac{4x^{2}-4x + 8}{x - 3}=+\infty$. As $x\to3^{-}$, the numerator $4x^{2}-4x + 8>0$ and the denominator $x - 3<0$ and approaches 0. So, $\lim_{x\to3^{-}}\frac{4x^{2}-4x + 8}{x - 3}=-\infty$. Since the left - hand limit and the right - hand limit are not equal, the two - sided limit does not exist.

Answer:

DNE