Math Problem Statement
найди точки разрыва и определи их характер
Solution
To find the points of discontinuity and determine their nature for the given piecewise function:
e^{\frac{1}{\arctan x}}, & x < 0 \\ \frac{\sqrt{1 - \sin\left(\frac{\pi^2}{2x}\right)}}{\pi - x}, & x > 0 \end{cases}$$ ### Step 1: Identify the point where the function may be discontinuous The function is defined in two distinct cases for $$x < 0$$ and $$x > 0$$. The potential point of discontinuity is at $$x = 0$$ because this is where the behavior of the function changes. ### Step 2: Examine the behavior of $$f(x)$$ as $$x \to 0^-$$ and $$x \to 0^+$$ 1. **For $$x < 0$$:** The expression is $$e^{\frac{1}{\arctan x}}$$. - As $$x \to 0^-$$, $$\arctan x \to 0$$. - Thus, $$\frac{1}{\arctan x} \to \infty$$, causing $$e^{\frac{1}{\arctan x}} \to \infty$$. - Therefore, as $$x \to 0^-$$, $$f(x)$$ approaches infinity, indicating a vertical asymptote on the left side of zero. 2. **For $$x > 0$$:** The expression is $$\frac{\sqrt{1 - \sin\left(\frac{\pi^2}{2x}\right)}}{\pi - x}$$. - As $$x \to 0^+$$, $$\frac{\pi^2}{2x} \to \infty$$. - The behavior of $$\sin\left(\frac{\pi^2}{2x}\right)$$ oscillates between -1 and 1 as $$x$$ approaches zero, causing the square root term to vary. - This oscillatory behavior combined with the $$\pi - x$$ term in the denominator causes the function to oscillate without settling to a limit as $$x \to 0^+$$. ### Conclusion Since the left-hand limit $$\lim_{x \to 0^-} f(x) = \infty$$ and the right-hand limit $$\lim_{x \to 0^+} f(x)$$ does not exist due to oscillations, we conclude: - There is a **discontinuity at $$x = 0$$**. - The nature of this discontinuity is an **infinite discontinuity** on the left and an **oscillatory discontinuity** on the right. Would you like further details on this analysis or additional explanation on the behavior of the function around $$x = 0$$? Here are five related questions to explore further: 1. How would you classify discontinuities in general for different types of functions? 2. What would the graph of this function look like around $$x = 0$$? 3. How does oscillatory behavior affect the limit properties of functions? 4. Can other points of discontinuity exist in piecewise functions like this one? 5. What role does the exponential function play in creating an infinite discontinuity? **Tip:** When analyzing discontinuities in piecewise functions, always check limits from both sides of the point of interest to determine the exact nature of the discontinuity.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity and Discontinuity
Limits
Oscillatory Behavior
Infinite Discontinuity
Formulas
f(x) = e^{1 / arctan(x)}, x < 0
f(x) = sqrt(1 - sin(pi^2 / (2x))) / (pi - x), x > 0
Theorems
Limit Definition of Continuity
Behavior of Sine and Exponential Functions
Suitable Grade Level
Undergraduate Level (Calculus)
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