IDNLearn.com: Your one-stop destination for finding reliable answers. Ask your questions and receive comprehensive, trustworthy responses from our dedicated team of experts.
Sagot :
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.
[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.
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Your questions find clarity at IDNLearn.com. Thanks for stopping by, and come back for more dependable solutions.
