Question

evaluate the limit, if it exists. (if an answer does not exist, enter dne.)
\lim_{x\to2}\frac{x^{2}-8x + 4}{x - 2}
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Explanation:

Step1: Check direct - substitution

If we substitute \(x = 2\) into \(\frac{x^{2}-8x + 4}{x - 2}\), we get \(\frac{2^{2}-8\times2 + 4}{2 - 2}=\frac{4-16 + 4}{0}=\frac{-8}{0}\), which is undefined. So, we need to factor or simplify the numerator.

Step2: Try to factor the numerator

The numerator \(x^{2}-8x + 4\) cannot be factored easily. Let's use the polynomial long - division or rewrite the numerator as follows:

$$
LATEXBLOCK0
$$

We can also use the fact that if \(\lim_{x
ightarrow a}\frac{f(x)}{x - a}\) and \(f(a)=0\), we can try to find the other factor of \(f(x)\) such that \(f(x)=(x - a)g(x)\).
Let's assume \(x^{2}-8x + 4=(x - 2)(x + b)+c\). Expanding \((x - 2)(x + b)+c=x^{2}+bx-2x-2b + c=x^{2}+(b - 2)x+(c - 2b)\).
Comparing coefficients: \(b-2=-8\Rightarrow b=-6\) and \(c-2b = 4\). Substituting \(b=-6\) into \(c-2b = 4\), we get \(c-2\times(-6)=4\Rightarrow c=4 - 12=-8\). So \(x^{2}-8x + 4=(x - 2)(x-6)-8\).
Then \(\frac{x^{2}-8x + 4}{x - 2}=\frac{(x - 2)(x - 6)-8}{x - 2}=x - 6-\frac{8}{x - 2}\).
As \(x
ightarrow2\), \(\lim_{x
ightarrow2}(x - 6-\frac{8}{x - 2})\) does not exist.

Answer:

DNE