Math Problem Statement
Evaluate the limits. f(x)=⎧⎩⎨⎪⎪5x2−65x3+65x2+6x<−1−1≤x≤1x>1 Enter DNE if the limit does not exist. a) limx→−1−f(x) =
b) limx→−1+f(x) =
c) limx→−1f(x) =
d) f(−1) =
e) limx→1−f(x) =
f) limx→1+f(x) =
g) limx→1f(x) =
h) f(1) =
Solution
We are given a piecewise function:
5x^2 - 6 & \text{if } x < -1 \\ 5x^3 + 6 & \text{if } -1 \leq x \leq 1 \\ x & \text{if } x > 1 \end{cases}$$ ### a) $$\lim_{x \to -1^-} f(x)$$ This is the left-hand limit as $$x$$ approaches $$-1$$. For $$x < -1$$, the function is defined as $$f(x) = 5x^2 - 6$$. We substitute $$x = -1$$ into this expression: $$\lim_{x \to -1^-} f(x) = 5(-1)^2 - 6 = 5(1) - 6 = -1$$ ### b) $$\lim_{x \to -1^+} f(x)$$ This is the right-hand limit as $$x$$ approaches $$-1$$. For $$-1 \leq x \leq 1$$, the function is defined as $$f(x) = 5x^3 + 6$$. We substitute $$x = -1$$ into this expression: $$\lim_{x \to -1^+} f(x) = 5(-1)^3 + 6 = 5(-1) + 6 = 1$$ ### c) $$\lim_{x \to -1} f(x)$$ Since the left-hand and right-hand limits are different ($$-1$$ and $$1$$, respectively), the limit does not exist at $$x = -1$$: $$\lim_{x \to -1} f(x) = \text{DNE}$$ ### d) $$f(-1)$$ The value of the function at $$x = -1$$ is found from the second piece, since $$-1 \leq x \leq 1$$ applies here. Thus, $$f(-1) = 5(-1)^3 + 6 = 5(-1) + 6 = 1$$ ### e) $$\lim_{x \to 1^-} f(x)$$ This is the left-hand limit as $$x$$ approaches $$1$$. For $$-1 \leq x \leq 1$$, the function is $$f(x) = 5x^3 + 6$$. We substitute $$x = 1$$ into this expression: $$\lim_{x \to 1^-} f(x) = 5(1)^3 + 6 = 5(1) + 6 = 11$$ ### f) $$\lim_{x \to 1^+} f(x)$$ This is the right-hand limit as $$x$$ approaches $$1$$. For $$x > 1$$, the function is $$f(x) = x$$. We substitute $$x = 1$$: $$\lim_{x \to 1^+} f(x) = 1$$ ### g) $$\lim_{x \to 1} f(x)$$ Since the left-hand and right-hand limits are different ($$11$$ and $$1$$, respectively), the limit does not exist at $$x = 1$$: $$\lim_{x \to 1} f(x) = \text{DNE}$$ ### h) $$f(1)$$ The value of the function at $$x = 1$$ is given by the second piece, where $$-1 \leq x \leq 1$$: $$f(1) = 5(1)^3 + 6 = 5(1) + 6 = 11$$ ### Final Answers: a) $$\lim_{x \to -1^-} f(x) = -1$$ b) $$\lim_{x \to -1^+} f(x) = 1$$ c) $$\lim_{x \to -1} f(x) = \text{DNE}$$ d) $$f(-1) = 1$$ e) $$\lim_{x \to 1^-} f(x) = 11$$ f) $$\lim_{x \to 1^+} f(x) = 1$$ g) $$\lim_{x \to 1} f(x) = \text{DNE}$$ h) $$f(1) = 11$$ Do you need further details or have any questions? --- Here are some related questions to expand your understanding: 1. What are the conditions for a limit to exist at a point? 2. How can a piecewise function have discontinuities at specific points? 3. Can the function still be continuous even if one of its piecewise components is a simple function (e.g., a polynomial)? 4. How would the limits change if the middle piece were a different polynomial? 5. What is the difference between jump discontinuity and removable discontinuity? **Tip:** When evaluating limits of piecewise functions, always consider the behavior from both sides of the point where the function definition changes.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Discontinuities
Formulas
Limit formulas for piecewise functions
Limit rules (left-hand limit, right-hand limit)
Theorems
Definition of Limits
Existence of Limits Theorem
Piecewise Function Limit Theorem
Suitable Grade Level
Grade 12 or First-Year College Calculus