Math Problem Statement
Найти точки разрыва функции f ( x ) и определить их характер.
Solution
To find the discontinuities of the function and determine their nature, let's analyze the piecewise function given:
e^{\frac{1}{\operatorname{arctg} \, x}}, & x < 0 \\ \frac{\sqrt{1 - \sin \frac{\pi^2}{2x}}}{\pi - x}, & x > 0 \end{cases}$$ Here's the approach we'll take to solve this problem: 1. **Identify Potential Points of Discontinuity:** Since this is a piecewise function with different definitions for $$x < 0$$ and $$x > 0$$, we should examine the behavior of $$f(x)$$ at $$x = 0$$ (the boundary between the two cases) as it’s a possible point of discontinuity. 2. **Examine Left and Right Limits at $$x = 0$$:** - **Left-hand limit ($$x \to 0^-$$):** Evaluate $$\lim_{x \to 0^-} e^{\frac{1}{\operatorname{arctg} \, x}}$$. - As $$x \to 0^-$$, $$\operatorname{arctg} \, x \to 0$$, so $$\frac{1}{\operatorname{arctg} \, x} \to \infty$$. - Consequently, $$e^{\frac{1}{\operatorname{arctg} \, x}} \to e^{\infty} = \infty$$. - This means the left-hand limit does not approach a finite value; instead, it tends to infinity. - **Right-hand limit ($$x \to 0^+$$):** Evaluate $$\lim_{x \to 0^+} \frac{\sqrt{1 - \sin \frac{\pi^2}{2x}}}{\pi - x}$$. - The behavior of $$\sin \frac{\pi^2}{2x}$$ as $$x \to 0^+$$ is oscillatory, leading to undefined or oscillating behavior in the square root and denominator. - This makes it difficult to approach a single finite limit from the right side as $$x \to 0^+$$. 3. **Conclusion on Discontinuity at $$x = 0$$:** Since both the left-hand and right-hand limits are not finite and do not match, there is a **discontinuity at $$x = 0$$**. Specifically, this discontinuity is an **essential (infinite) discontinuity** because the limits do not converge to any finite value. In summary, the function $$f(x)$$ has a discontinuity at $$x = 0$$, and it is an essential discontinuity due to divergent behavior from both sides. Would you like further details on any step or a deeper exploration of the function's properties? --- ### Related Questions: 1. What is an essential discontinuity and how does it differ from a removable or jump discontinuity? 2. How does the oscillatory nature of $$\sin \frac{\pi^2}{2x}$$ affect the limit as $$x \to 0^+$$? 3. Could we use series expansions to approximate the behavior of $$f(x)$$ around $$x = 0$$? 4. How would the graph of this function look near $$x = 0$$? 5. What would happen if we defined $$f(0)$$ in some way to attempt to remove the discontinuity? ### Tip: When working with piecewise functions, always check boundary points between cases, as these are often where discontinuities occur.Ask a new question for Free
By Image
Drop file here or Click Here to upload
Math Problem Analysis
Mathematical Concepts
Calculus
Limits
Discontinuity
Piecewise Functions
Oscillatory Behavior
Formulas
e^{1 / \operatorname{arctg} \, x}, x < 0
\frac{\sqrt{1 - \sin \frac{\pi^2}{2x}}}{\pi - x}, x > 0
Theorems
Limit Theorem
Continuity and Discontinuity Theorem
Suitable Grade Level
Undergraduate Calculus
Related Recommendation
Determine the Type of Discontinuity at x=0 for a Piecewise Function
Discussing the Continuity of a Piecewise Function f(x)
Is the Piecewise Function f(x) Continuous? A Step-by-Step Solution
Understanding Piecewise Functions and Discontinuities at x = 1.5
Limit Analysis of a Piecewise Function with Potential Discontinuities