Math Problem Statement
Determine if there exist values of a, b ∈ ℝ, such that the limits lim(x → 0) f(x) and lim(x → 3) f(x) exist for the given piecewise function.
Solution
We are given a piecewise function , and we are asked to determine if there exist real values for and such that the limits and exist.
The function is defined as:
\begin{cases} 2x + a & \text{if } x < 0 \\ \frac{4 - \sqrt{4x + 4}}{x^2 - 2x - 3} & \text{if } x \geq 0 \text{ and } x \neq 3 \\ b & \text{if } x = 3 \end{cases}$$ ### Step 1: Analyze the limit as $$x \to 0$$ We need to calculate: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2x + a) = a$$ and $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{4 - \sqrt{4x + 4}}{x^2 - 2x - 3}$$ First, simplify the expression for $$x \geq 0$$: $$4 - \sqrt{4x + 4} = 4 - 2\sqrt{x + 1}$$ Thus, the expression becomes: $$f(x) = \frac{4 - 2\sqrt{x + 1}}{(x - 3)(x + 1)}$$ Now, compute the limit: $$\lim_{x \to 0^+} \frac{4 - 2\sqrt{x + 1}}{(x - 3)(x + 1)}$$ For $$x \to 0$$, this expression evaluates as: $$\frac{4 - 2\sqrt{1}}{(-3)(1)} = \frac{4 - 2}{-3} = \frac{2}{-3} = -\frac{2}{3}$$ ### Condition for the limit at $$x = 0$$ to exist: $$a = -\frac{2}{3}$$ ### Step 2: Analyze the limit as $$x \to 3$$ Next, we evaluate the limits around $$x = 3$$: $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} \frac{4 - 2\sqrt{x + 1}}{(x - 3)(x + 1)}$$ $$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{4 - 2\sqrt{x + 1}}{(x - 3)(x + 1)}$$ Both limits involve a 0/0 indeterminate form, so we apply L'Hopital's rule: 1. Differentiate the numerator: $$\frac{d}{dx}[4 - 2\sqrt{x + 1}] = \frac{-1}{\sqrt{x + 1}}$$ 2. Differentiate the denominator: $$\frac{d}{dx}[(x - 3)(x + 1)] = (x - 3) + (x + 1) = 2x - 2$$ Now apply L'Hopital's rule at $$x = 3$$: $$\lim_{x \to 3} \frac{\frac{-1}{\sqrt{x + 1}}}{2x - 2} = \frac{-1/\sqrt{4}}{6 - 2} = \frac{-1/2}{4} = -\frac{1}{8}$$ ### Condition for the limit at $$x = 3$$ to exist: $$b = -\frac{1}{8}$$ ### Conclusion: The values of $$a$$ and $$b$$ that ensure the existence of the limits are: $$a = -\frac{2}{3}, \quad b = -\frac{1}{8}$$ Would you like further details or have any questions? --- Here are 5 related questions to deepen your understanding: 1. What is the importance of ensuring the limits match from both sides at a point of discontinuity? 2. How does L'Hopital's Rule help in resolving indeterminate forms like 0/0? 3. What would happen if the function had different expressions for $$x = 3$$ on both sides? 4. Can you find a case where the piecewise function has a jump discontinuity? 5. How do you compute higher-order limits at points of discontinuity? **Tip:** Always check the behavior of the function from both sides when evaluating limits, especially for piecewise functions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Continuity
L'Hopital's Rule
Formulas
lim(x → 0⁻) f(x) = a
lim(x → 0⁺) f(x) = -2/3
L'Hopital's Rule: lim(x → 3) f(x) using derivatives
Theorems
L'Hopital's Rule
Continuity at a Point
Suitable Grade Level
Undergraduate Calculus