Math Problem Statement
Diberikan fungsi h, dengan h(x) = { x^2 + (1 - cos x) / (2x sin 2x), x > 0; a / (x - 2), x ≤ 0 }. Tentukan nilai a agar lim x→0 h(x) ada.
Solution
The function is given as:
x^2 + \frac{1 - \cos x}{2x \sin 2x}, & x > 0 \\ \frac{a}{x - 2}, & x \leq 0 \end{cases}$$ We are asked to find the value of $$a$$ such that $$\lim_{x \to 0} h(x)$$ exists. To solve this, we need to check the limits of both pieces of the function as $$x \to 0$$ from the left ($$x \leq 0$$) and from the right ($$x > 0$$). ### 1. Right-hand limit ($$x \to 0^+$$): For $$x > 0$$, the function is: $$h(x) = x^2 + \frac{1 - \cos x}{2x \sin 2x}$$ As $$x \to 0^+$$, the first term $$x^2 \to 0$$. For the second term, apply L'Hopital's Rule, as it is of the indeterminate form $$\frac{0}{0}$$ when $$x \to 0$$: $$\lim_{x \to 0^+} \frac{1 - \cos x}{2x \sin 2x}$$ Differentiate the numerator and denominator with respect to $$x$$: - Numerator: $$\frac{d}{dx}(1 - \cos x) = \sin x$$ - Denominator: $$\frac{d}{dx}(2x \sin 2x) = 2\sin 2x + 4x \cos 2x$$ Thus, applying L'Hopital's Rule: $$\lim_{x \to 0^+} \frac{\sin x}{2\sin 2x + 4x \cos 2x} = \frac{0}{0 + 0} = 0$$ So, the right-hand limit is: $$\lim_{x \to 0^+} h(x) = 0 + 0 = 0$$ ### 2. Left-hand limit ($$x \to 0^-$$): For $$x \leq 0$$, the function is: $$h(x) = \frac{a}{x - 2}$$ As $$x \to 0^-$$, the left-hand limit becomes: $$\lim_{x \to 0^-} \frac{a}{x - 2} = \frac{a}{-2}$$ ### 3. Equating limits: For the limit to exist at $$x = 0$$, the right-hand limit must equal the left-hand limit. Therefore, we equate the two: $$0 = \frac{a}{-2}$$ Solving for $$a$$: $$a = 0$$ Thus, the value of $$a$$ such that $$\lim_{x \to 0} h(x)$$ exists is $$\boxed{0}$$. Would you like more details or explanations on any part of the solution? Here are 5 related questions for practice: 1. What is L'Hopital's Rule, and when can it be applied? 2. How do you calculate the limit of trigonometric functions as $$x \to 0$$? 3. What are the conditions for the existence of a limit at a point? 4. Can the limit of a piecewise function be discontinuous at a point? 5. How do you solve indeterminate forms of type $$\frac{0}{0}$$ using derivatives? **Tip:** When solving limits involving trigonometric functions, always check for indeterminate forms and consider using L'Hopital’s Rule if applicable.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Piecewise Functions
Trigonometric Limits
Formulas
L'Hopital's Rule
Trigonometric Limit: lim x→0 (1 - cos x) / x = 0
Theorems
Limit Theorem for Piecewise Functions
L'Hopital's Rule
Suitable Grade Level
Grades 11-12 or Undergraduate
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