To find the limit of the given expression as [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex], we need to carefully examine the function:
[tex]\[
\lim_{{x \to -2}} \frac{x^2 - 5x + 4}{x^2 - 2x - 8}
\][/tex]
Step-by-step, we proceed as follows:
1. Factorize both the numerator and the denominator:
- For the numerator [tex]\(x^2 - 5x + 4\)[/tex], we need to find the factors of 4 that add up to -5. These factors are -1 and -4, so we can write:
[tex]\[
x^2 - 5x + 4 = (x - 1)(x - 4)
\][/tex]
- For the denominator [tex]\(x^2 - 2x - 8\)[/tex], we need to find the factors of -8 that add up to -2. These factors are -4 and 2, so we can write:
[tex]\[
x^2 - 2x - 8 = (x - 4)(x + 2)
\][/tex]
2. Rewrite the function with the factored expressions:
[tex]\[
\frac{x^2 - 5x + 4}{x^2 - 2x - 8} = \frac{(x - 1)(x - 4)}{(x - 4)(x + 2)}
\][/tex]
3. Simplify the expression by canceling out the common factor [tex]\((x - 4)\)[/tex]:
[tex]\[
\frac{(x - 1)(x - 4)}{(x - 4)(x + 2)} = \frac{x - 1}{x + 2} \quad \text{for} \quad x \neq 4
\][/tex]
4. Evaluate the limit of the simplified expression as [tex]\( x \)[/tex] approaches [tex]\(-2\)[/tex]:
[tex]\[
\lim_{{x \to -2}} \frac{x - 1}{x + 2}
\][/tex]
Substituting [tex]\( x = -2 \)[/tex] into the simplified expression:
[tex]\[
\frac{(-2) - 1}{(-2) + 2} = \frac{-3}{0}
\][/tex]
Since [tex]\(\frac{-3}{0}\)[/tex] indicates division by zero, we need to consider the behavior around [tex]\(x = -2\)[/tex].
5. Examine the nature of the discontinuity:
- If we approach [tex]\(-2\)[/tex] from the left, the denominator is slightly negative and as we approach [tex]\(-2\)[/tex] from the right, the denominator is slightly positive. This leads to the function approaching [tex]\(-\infty\)[/tex] or [tex]\(\infty\)[/tex].
Hence:
[tex]\[
\lim_{{x \to -2}} \frac{x - 1}{x + 2} \quad \text{does not exist}
\][/tex]
The limit does not exist because the function approaches [tex]\(-\infty\)[/tex] or [tex]\(\infty\)[/tex] as x approaches [tex]\(-2\)[/tex] from the left and right respectively.
