Math Problem Statement
The function f is continuous at x = 0 and is defined for -1 < x < 1 by: f(x) = { (2a/x)(e^x - 1) if -1 < x < 0, 1 if x = 0, (bx cos(x)) / (1 - sqrt(1 - x)) if 0 < x < 1 }. Determine the values of the constants a and b.
Solution
We are given that the function is continuous at . The function is defined in three parts:
\begin{cases} \frac{2a}{x}(e^x - 1), & \text{if } -1 < x < 0, \\ 1, & \text{if } x = 0, \\ \frac{b x \cos x}{1 - \sqrt{1 - x}}, & \text{if } 0 < x < 1. \end{cases}$$ To determine the values of constants $$a$$ and $$b$$, we will use the condition of continuity at $$x = 0$$. This requires that the left-hand limit as $$x \to 0^-$$, the value of the function at $$x = 0$$, and the right-hand limit as $$x \to 0^+$$ must all be equal. That is: $$\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x).$$ ### Step 1: Continuity at $$x = 0$$ from the left-hand side Consider the left-hand limit as $$x \to 0^-$$: $$f(x) = \frac{2a}{x}(e^x - 1).$$ Using the fact that $$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$, we have: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{2a}{x}(e^x - 1) = 2a.$$ ### Step 2: Continuity at $$x = 0$$ We are given that $$f(0) = 1$$. Therefore, from the condition of continuity: $$2a = 1.$$ Solving for $$a$$: $$a = \frac{1}{2}.$$ ### Step 3: Continuity at $$x = 0$$ from the right-hand side Now consider the right-hand limit as $$x \to 0^+$$: $$f(x) = \frac{b x \cos x}{1 - \sqrt{1 - x}}.$$ We need to evaluate this limit as $$x \to 0^+$$. First, use a series expansion for the square root term: $$\sqrt{1 - x} \approx 1 - \frac{x}{2} \text{ as } x \to 0.$$ Thus, $$1 - \sqrt{1 - x} \approx \frac{x}{2} \text{ as } x \to 0.$$ Now, we substitute this into the expression for $$f(x)$$ for $$x > 0$$: $$f(x) \approx \frac{b x \cos x}{\frac{x}{2}} = 2b \cos x.$$ As $$x \to 0^+$$, $$\cos x \to 1$$, so: $$\lim_{x \to 0^+} f(x) = 2b.$$ ### Step 4: Continuity condition Since $$f(0) = 1$$, we also have from the right-hand limit: $$2b = 1.$$ Solving for $$b$$: $$b = \frac{1}{2}.$$ ### Conclusion: The values of the constants are: $$a = \frac{1}{2}, \quad b = \frac{1}{2}.$$ Would you like further details on any steps? Here are 5 related questions to consider: 1. How do you compute limits involving series expansions for functions like $$e^x$$ or $$\sqrt{1-x}$$? 2. What is the significance of a function being continuous at a point? 3. How would the solution change if the function was required to be differentiable at $$x = 0$$? 4. What are some other common techniques for determining limits of complex functions? 5. How can you use the definition of the derivative to evaluate limits? **Tip**: Always check if you can simplify a complex expression using series expansions or small approximations when evaluating limits.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Piecewise Functions
Series Expansion
Formulas
Limit definition for continuity: lim_{x → 0^-} f(x) = f(0) = lim_{x → 0^+} f(x)
Exponential limit: lim_{x → 0} (e^x - 1) / x = 1
Square root expansion: sqrt(1 - x) ≈ 1 - x/2 as x → 0
Theorems
Limit Theorem
Continuity at a point
Suitable Grade Level
College-Level Calculus