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

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

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

Answer

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: [ \begin{align*} x^{2}-8x + 4&=x^{2}-2x-6x + 4\ &=x(x - 2)-6x+4 \end{align*} ] We can also use the fact that if (\lim_{x\rightarrow 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\rightarrow2), (\lim_{x\rightarrow2}(x - 6-\frac{8}{x - 2})) does not exist.

Answer:

DNE