Math Problem Statement
Tính f(1); lim x → 1+ f(x); lim x → 1- f(x) for the piecewise function f(x) defined as tan(√(1 - x)) for x ≤ 1, and sin²(√(x² - 1)) / (3x² - 3) for x > 1.
Solution
The function is defined as a piecewise function:
\begin{cases} \text{tan}\sqrt{1-x}, & \text{for } x \leq 1 \\ \frac{\sin^2 \sqrt{x^2 - 1}}{3x^2 - 3}, & \text{for } x > 1 \end{cases}$$ You are asked to find: 1. $$f(1)$$ 2. $$\lim_{x \to 1^+} f(x)$$ 3. $$\lim_{x \to 1^-} f(x)$$ ### 1. Finding $$f(1)$$: Since $$f(x) = \text{tan}\sqrt{1-x}$$ for $$x \leq 1$$, we use this expression to evaluate $$f(1)$$: $$f(1) = \text{tan}\sqrt{1 - 1} = \text{tan}(0) = 0$$ Thus, $$f(1) = 0$$. ### 2. Finding $$\lim_{x \to 1^+} f(x)$$: For $$x > 1$$, the function is: $$f(x) = \frac{\sin^2 \sqrt{x^2 - 1}}{3x^2 - 3}$$ We need to compute the limit as $$x \to 1^+$$: As $$x \to 1^+$$, we observe that: $$\sqrt{x^2 - 1} \to \sqrt{1^2 - 1} = \sqrt{0} = 0$$ and $$3x^2 - 3 = 3(1^2) - 3 = 3 - 3 = 0$$ Thus, we are dealing with a $$\frac{0}{0}$$ indeterminate form. To resolve this, we can apply L'Hopital's Rule by differentiating the numerator and denominator with respect to $$x$$. #### Applying L'Hopital's Rule: The numerator is $$\sin^2(\sqrt{x^2 - 1})$$, and the denominator is $$3x^2 - 3$$. Differentiating both: - Derivative of the numerator $$\sin^2(\sqrt{x^2 - 1})$$: $$\frac{d}{dx} \sin^2(\sqrt{x^2 - 1}) = 2 \sin(\sqrt{x^2 - 1}) \cdot \cos(\sqrt{x^2 - 1}) \cdot \frac{d}{dx}(\sqrt{x^2 - 1}) = 2 \sin(\sqrt{x^2 - 1}) \cos(\sqrt{x^2 - 1}) \cdot \frac{x}{\sqrt{x^2 - 1}}$$ At $$x = 1^+$$, this expression approaches 0 because $$\sin(\sqrt{x^2 - 1})$$ approaches 0. - Derivative of the denominator $$3x^2 - 3$$: $$\frac{d}{dx}(3x^2 - 3) = 6x$$ At $$x = 1^+$$, this equals 6. Thus, applying L'Hopital's Rule: $$\lim_{x \to 1^+} f(x) = \frac{0}{6} = 0$$ ### 3. Finding $$\lim_{x \to 1^-} f(x)$$: For $$x \leq 1$$, the function is $$f(x) = \text{tan}(\sqrt{1 - x})$$. As $$x \to 1^-$$: $$\sqrt{1 - x} \to \sqrt{0} = 0$$ and $$\text{tan}(0) = 0$$ Thus, $$\lim_{x \to 1^-} f(x) = 0$$. ### Conclusion: - $$f(1) = 0$$ - $$\lim_{x \to 1^+} f(x) = 0$$ - $$\lim_{x \to 1^-} f(x) = 0$$ Since the left-hand limit, right-hand limit, and the value of the function at $$x = 1$$ are all equal, the function is continuous at $$x = 1$$. --- Would you like to explore more details or ask any further questions? Here are five related questions you might find interesting: 1. How is L'Hopital's Rule applied to other indeterminate forms like $$\frac{\infty}{\infty}$$? 2. Can we use different methods, such as series expansion, to solve the limit problem? 3. How does the continuity of a piecewise function get affected by different kinds of discontinuities? 4. What happens if the limits on both sides of $$x = 1$$ were different? 5. How do we determine higher-order differentiability at points where a function is piecewise? **Tip**: Always check the differentiability of functions when using L'Hopital's Rule, especially at critical points like $$x = 1$$.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Limits
Continuity
Piecewise Functions
L'Hopital's Rule
Formulas
tan(√(1 - x)) for x ≤ 1
sin²(√(x² - 1)) / (3x² - 3) for x > 1
L'Hopital's Rule for indeterminate forms
Theorems
L'Hopital's Rule
Continuity of Piecewise Functions
Suitable Grade Level
Grades 11-12 or Early College
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