Answer :
Sure, let's evaluate the given piecewise function step-by-step.
Given the function:
[tex]\[ f(x) = \begin{cases} -5x & \text{if } x < 1 \\ 1 & \text{if } x = 1 \\ 5x & \text{if } x > 1 \end{cases} \][/tex]
(a) [tex]\(\lim_{x \to 1^-} f(x)\)[/tex]
- This represents the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 1 from the left (i.e., from values less than 1).
- When [tex]\(x < 1\)[/tex], [tex]\(f(x) = -5x\)[/tex].
Thus,
[tex]\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} -5x = -5 \cdot 1 = -5 \][/tex]
(b) [tex]\(\lim_{x \to 1^+} f(x)\)[/tex]
- This represents the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 1 from the right (i.e., from values greater than 1).
- When [tex]\(x > 1\)[/tex], [tex]\(f(x) = 5x\)[/tex].
Thus,
[tex]\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5x = 5 \cdot 1 = 5 \][/tex]
(c) [tex]\(\lim_{x \to 1} f(x)\)[/tex]
- For the limit [tex]\(\lim_{x \to 1} f(x)\)[/tex] to exist, both the left-hand limit [tex]\(\lim_{x \to 1^-} f(x)\)[/tex] and the right-hand limit [tex]\(\lim_{x \to 1^+} f(x)\)[/tex] must exist and be equal.
- From (a), we have [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex].
- From (b), we have [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex].
Since these two limits are not equal, the overall limit does not exist (DNE).
(d) [tex]\(f(1)\)[/tex]
- The function value at [tex]\(x = 1\)[/tex] is given by the piecewise definition. It specifically states that [tex]\(f(x) = 1\)[/tex] when [tex]\(x = 1\)[/tex].
Thus,
[tex]\[ f(1) = 1 \][/tex]
(e) Comparison and conclusion
- Since the left-hand limit [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex] and the right-hand limit [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex] are not equal, there is a discontinuity at [tex]\(x = 1\)[/tex].
- The type of discontinuity where the left-hand and right-hand limits exist but are not equal is called a "jump discontinuity".
So, the function has a jump discontinuity at [tex]\(x = 1\)[/tex].
### Summary of Results:
(a) [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex]
(b) [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex]
(c) [tex]\(\lim_{x \to 1} f(x) = \text{DNE (Does Not Exist)}\)[/tex]
(d) [tex]\(f(1) = 1\)[/tex]
(e) The function has a jump discontinuity at [tex]\(x = 1\)[/tex].
Given the function:
[tex]\[ f(x) = \begin{cases} -5x & \text{if } x < 1 \\ 1 & \text{if } x = 1 \\ 5x & \text{if } x > 1 \end{cases} \][/tex]
(a) [tex]\(\lim_{x \to 1^-} f(x)\)[/tex]
- This represents the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 1 from the left (i.e., from values less than 1).
- When [tex]\(x < 1\)[/tex], [tex]\(f(x) = -5x\)[/tex].
Thus,
[tex]\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} -5x = -5 \cdot 1 = -5 \][/tex]
(b) [tex]\(\lim_{x \to 1^+} f(x)\)[/tex]
- This represents the limit of [tex]\(f(x)\)[/tex] as [tex]\(x\)[/tex] approaches 1 from the right (i.e., from values greater than 1).
- When [tex]\(x > 1\)[/tex], [tex]\(f(x) = 5x\)[/tex].
Thus,
[tex]\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5x = 5 \cdot 1 = 5 \][/tex]
(c) [tex]\(\lim_{x \to 1} f(x)\)[/tex]
- For the limit [tex]\(\lim_{x \to 1} f(x)\)[/tex] to exist, both the left-hand limit [tex]\(\lim_{x \to 1^-} f(x)\)[/tex] and the right-hand limit [tex]\(\lim_{x \to 1^+} f(x)\)[/tex] must exist and be equal.
- From (a), we have [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex].
- From (b), we have [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex].
Since these two limits are not equal, the overall limit does not exist (DNE).
(d) [tex]\(f(1)\)[/tex]
- The function value at [tex]\(x = 1\)[/tex] is given by the piecewise definition. It specifically states that [tex]\(f(x) = 1\)[/tex] when [tex]\(x = 1\)[/tex].
Thus,
[tex]\[ f(1) = 1 \][/tex]
(e) Comparison and conclusion
- Since the left-hand limit [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex] and the right-hand limit [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex] are not equal, there is a discontinuity at [tex]\(x = 1\)[/tex].
- The type of discontinuity where the left-hand and right-hand limits exist but are not equal is called a "jump discontinuity".
So, the function has a jump discontinuity at [tex]\(x = 1\)[/tex].
### Summary of Results:
(a) [tex]\(\lim_{x \to 1^-} f(x) = -5\)[/tex]
(b) [tex]\(\lim_{x \to 1^+} f(x) = 5\)[/tex]
(c) [tex]\(\lim_{x \to 1} f(x) = \text{DNE (Does Not Exist)}\)[/tex]
(d) [tex]\(f(1) = 1\)[/tex]
(e) The function has a jump discontinuity at [tex]\(x = 1\)[/tex].
