$$ \left. \begin{array} { l } { \lim _ { x ightarrow 2 } \frac { x - 2 } { 2 x ^ { 2 } - 8 x + 8 } } \ { \vdots } \ { \vdots } \end{array} ight. $$

Question

$$ \left. \begin{array} { l } { \lim _ { x ightarrow 2 } \frac { x - 2 } { 2 x ^ { 2 } - 8 x + 8 } } \ { \vdots } \ { \vdots } \end{array} ight. $$

UpStudy AI · Accepted Answer

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.
The limit does not exist because it approaches infinity as $$ x $$ approaches 2.

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