\( \left. \begin{array} { l } { \lim _ { x \rightarrow 2 } \frac { x - 2 } { 2 x ^ { 2 } - 8 x + 8 } } \\ { \vdots } \\ { \vdots } \end{array} \right. \)

Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow 2}\left(\frac{x-2}{2x^{2}-8x+8}\right)\) - step1: Reduce the fraction: \(\lim _{x\rightarrow 2}\left(\frac{1}{\left(x-2\right)\times 2}\right)\) - step2: Calculate: \(\textrm{The limit does not exist}\) To evaluate the limit \[ \lim_{x \rightarrow 2} \frac{x - 2}{2x^{2} - 8x + 8}, \] we first substitute \( x = 2 \) into the expression: 1. **Substituting \( x = 2 \)**: - The numerator becomes \( 2 - 2 = 0 \). - The denominator becomes \( 2(2^2) - 8(2) + 8 = 8 - 16 + 8 = 0 \). Since both the numerator and denominator approach 0, we have an indeterminate form \( \frac{0}{0} \). To resolve this, we can factor the denominator. 2. **Factoring the denominator**: - The expression \( 2x^{2} - 8x + 8 \) can be factored as follows: \[ 2(x^{2} - 4x + 4) = 2(x - 2)^{2}. \] 3. **Rewriting the limit**: Now we can rewrite the limit: \[ \lim_{x \rightarrow 2} \frac{x - 2}{2(x - 2)^{2}} = \lim_{x \rightarrow 2} \frac{1}{2(x - 2)}. \] 4. **Evaluating the limit**: As \( x \) approaches 2, \( (x - 2) \) approaches 0, which means \( \frac{1}{2(x - 2)} \) approaches infinity. Therefore, the limit does not exist. Thus, the final conclusion is that the limit does not exist.