Math Problem Statement
Diberikan fungsi h, dengan h(x) = { x^2 + (1 - cos x) / (2x sin 2x) untuk x > 0, a / (x - 2) untuk x <= 0. Tentukan nilai a agar lim (x -> 0) h(x) ada.
Solution
The problem provided asks to determine the value of so that the limit exists for the given piecewise function:
\begin{cases} x^2 + \frac{1 - \cos x}{2x \sin 2x} & x > 0 \\ \frac{a}{x - 2} & x \leq 0 \end{cases}$$ ### Step 1: Evaluate the limit from the right-hand side ($$x \to 0^+$$). For $$x > 0$$, we have: $$h(x) = x^2 + \frac{1 - \cos x}{2x \sin 2x}$$ - As $$x \to 0^+$$, $$x^2 \to 0$$. - We need to analyze the second term: $$\frac{1 - \cos x}{2x \sin 2x}$$. Using standard limits: - $$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$$, - $$\lim_{x \to 0} \frac{\sin 2x}{2x} = 1$$. Thus, we can evaluate the second term: $$\lim_{x \to 0} \frac{1 - \cos x}{2x \sin 2x} = \lim_{x \to 0} \frac{\frac{1 - \cos x}{x^2}}{\frac{2x \sin 2x}{x^2}} = 0$$ Therefore, the right-hand limit becomes: $$\lim_{x \to 0^+} h(x) = 0 + 0 = 0.$$ ### Step 2: Evaluate the limit from the left-hand side ($$x \to 0^-$$). For $$x \leq 0$$, the function is $$h(x) = \frac{a}{x - 2}$$. As $$x \to 0^-$$, we substitute directly into the expression: $$\lim_{x \to 0^-} h(x) = \frac{a}{0 - 2} = \frac{a}{-2} = -\frac{a}{2}.$$ ### Step 3: Set the limits equal to ensure continuity. For the limit $$\lim_{x \to 0} h(x)$$ to exist, the left-hand and right-hand limits must be equal: $$0 = -\frac{a}{2}.$$ Solving for $$a$$: $$a = 0.$$ Thus, the value of $$a$$ that ensures the limit exists is $$a = 0$$. --- Would you like further details or explanations? Here are five related questions for practice: 1. What is the general condition for a piecewise function to have a limit at a point? 2. How does L'Hôpital's rule help in solving limits involving indeterminate forms? 3. Why does $$\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$$? 4. How do you evaluate the limit of trigonometric functions as $$x \to 0$$? 5. Can a function be continuous but not differentiable at a point? **Tip:** When solving piecewise functions involving limits, always check both sides independently before equating them.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Trigonometric Limits
Formulas
lim (x -> 0) (1 - cos x) / x^2 = 1/2
lim (x -> 0) sin 2x / 2x = 1
Theorems
Limit Theorem
Continuity
Suitable Grade Level
Grades 11-12
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