Math Problem Statement
If function defined by F(x) = { (sin3x) / (2x) + (k + 1), for x ≠ 0; k, for x = 0 } is continuous at x = 0, then k equals:
Solution
The function provided is:
\begin{cases} \frac{\sin(3x)}{2x} + (k + 1), & x \neq 0 \\ k, & x = 0 \end{cases}$$ The function is continuous at $$x = 0$$, and we need to determine the value of $$k$$. ### Steps to solve: 1. **Continuity Condition**: For $$F(x)$$ to be continuous at $$x = 0$$, the left-hand limit, right-hand limit, and the value of the function at $$x = 0$$ must be equal. This means: $$\lim_{x \to 0} F(x) = F(0)$$ 2. **Evaluate the limit**: For $$x \neq 0$$, we have the function: $$F(x) = \frac{\sin(3x)}{2x} + (k + 1)$$ We need to find: $$\lim_{x \to 0} \left( \frac{\sin(3x)}{2x} + (k + 1) \right)$$ Using the standard limit result $$\lim_{x \to 0} \frac{\sin(ax)}{x} = a$$, we can calculate: $$\lim_{x \to 0} \frac{\sin(3x)}{2x} = \frac{3}{2}$$ So: $$\lim_{x \to 0} F(x) = \frac{3}{2} + (k + 1)$$ 3. **Continuity condition at $$x = 0$$**: $$\frac{3}{2} + (k + 1) = k$$ 4. **Solve for $$k$$**: $$\frac{3}{2} + k + 1 = k$$ Simplifying: $$\frac{3}{2} + 1 = 0$$ $$\frac{5}{2} = 0$$ This leads to an inconsistency since the equation cannot be true, indicating a different approachAsk a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Limits
Continuity
Formulas
Limit as x approaches 0 of sin(ax)/x = a
For continuity at a point x = c, lim(x→c) f(x) = f(c)
Theorems
Limit Theorem
Continuity Condition
Suitable Grade Level
Grades 11-12