Math Problem Statement
Find the value(s) of the constant a such that lim(x→1) f(x) exists, where f(x) is defined piecewise: f(x) = {ax^2 + 2x for x ≤ 1, (x^2 - 1)/(x - 1) for x > 1}.
Solution
The problem asks to find the value of the constant such that the limit of the function exists as approaches 1. The function is defined piecewise:
\begin{cases} ax^2 + 2x, & \text{if } x \leq 1 \\ \frac{x^2 - 1}{x - 1}, & \text{if } x > 1 \end{cases}$$ To ensure that the limit exists at $$x = 1$$, we need to make sure the left-hand limit ($$\lim_{x \to 1^-} f(x)$$) and the right-hand limit ($$\lim_{x \to 1^+} f(x)$$) are equal, and that they also match $$f(1)$$. **Step 1: Calculate the left-hand limit ($$\lim_{x \to 1^-} f(x)$$)** Since $$x \leq 1$$, we use the first piece of the function: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1} (ax^2 + 2x) = a(1)^2 + 2(1) = a + 2$$ **Step 2: Calculate the right-hand limit ($$\lim_{x \to 1^+} f(x)$$)** For $$x > 1$$, the function simplifies to: $$f(x) = \frac{x^2 - 1}{x - 1}$$ This expression can be simplified as follows: $$f(x) = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \text{ for } x \neq 1$$ Thus: $$\lim_{x \to 1^+} f(x) = \lim_{x \to 1} (x + 1) = 1 + 1 = 2$$ **Step 3: Equating limits to find $$a$$** For the limit to exist at $$x = 1$$, the left-hand and right-hand limits must be equal: $$a + 2 = 2$$ Solving for $$a$$: $$a = 0$$ Therefore, the value of $$a$$ that makes the limit exist as $$x \to 1$$ is $$\boxed{0}$$. Would you like more details on any of these steps, or do you have any other questions? Here are some related questions: 1. What happens if we change the function's definition for $$x \leq 1$$? 2. How do we handle continuity if the function has multiple points of interest? 3. What if the limit involves infinity? How do we handle such cases? 4. Can we have a limit if the function is undefined at the point of interest? 5. How do we approach limits involving trigonometric or exponential functions? **Tip:** When working with limits, always check if simplifying expressions can help, especially when dealing with piecewise functions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Algebra
Formulas
lim(x→1) f(x) for both the left-hand and right-hand limits
Quadratic expressions ax^2 + 2x
Simplified form of (x^2 - 1)/(x - 1) = x + 1 for x ≠ 1
Theorems
Limit definition
Continuity at a point
Suitable Grade Level
Grades 10-12 or early college
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