We are given that $-2 \leq f(x) \leq x^2 - 4x + 2$ for all $x \neq 2$, and we need to determine which of the statements must be true. Let's analyze each option.

Option 1: $\lim_{x \to 2} f(x) = -2$

• We know that $f(x)$ is bounded between $-2$ and $x^2 - 4x + 2$, but no additional information about the behavior of $f(x)$ as $x$ approaches 2 is given.
• Without more details about the behavior of $f(x)$ near $x = 2$, this limit does not have to equal $-2$. The upper bound $x^2 - 4x + 2$ evaluated at $x = 2$ is also $-2$, but this alone is not sufficient to conclude the limit equals $-2$.

Thus, Option 1 is not necessarily true.

Option 2: $f(2) = -2$

• The condition $f(x)$ is defined for all $x \neq 2$, and we are not given the value of $f(x)$ at $x = 2$. Therefore, there is no information to confirm that $f(2) = -2$.

Thus, Option 2 is not necessarily true because we don't know if $f(x)$ is even defined at $x = 2$.

Option 3: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)$

• The function $f(x)$ is continuous for all $x \neq 2$, and there is no discontinuity mentioned at $x = 0$.
• Since $f(x)$ is bounded by a continuous function $x^2 - 4x + 2$, the left-hand limit ($\lim_{x \to 0^-} f(x)$) and right-hand limit ($\lim_{x \to 0^+} f(x)$) must agree at $x = 0$, as there is no indication of a discontinuity at this point.

Thus, Option 3 is true.

Option 4: $\lim_{x \to 0} f(x)$ does not exist

• As discussed in Option 3, since $f(x)$ is bounded by a continuous function and there is no indication of a discontinuity at $x = 0$, the limit $\lim_{x \to 0} f(x)$ must exist.

Thus, Option 4 is false.

Conclusion:

The correct answer is:

• Option 3: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)$

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