Math Problem Statement
Consider the modified Dirichlet function $F(x)=\begin{cases} x, & x\notin \Q\ 0, & x\in \Q \end{cases}$. \begin{itemize} \item[(a)] Show that $\lim_{x \to 0} F(x)=0$. \item[(b)] Show that if $a\neq 0$, then $\lim_{x \to a} F(x)$ does not exist. \end{itemize}
Solution
Let's analyze the modified Dirichlet function:
x, & x \notin \mathbb{Q} \\ 0, & x \in \mathbb{Q} \end{cases}$$ --- ### **(a) Show that** $$\lim_{x \to 0} F(x) = 0$$. **Proof:** We need to verify that: $$\forall \epsilon > 0, \exists \delta > 0 \text{ such that if } |x - 0| < \delta, \text{ then } |F(x) - 0| < \epsilon.$$ 1. **For irrational points** $$(x \notin \mathbb{Q})$$: If $$x$$ is irrational, then $$F(x) = x$$, so: $$|F(x) - 0| = |x|.$$ For any $$\epsilon > 0$$, choose $$\delta = \epsilon$$. If $$|x| < \delta$$, then: $$|F(x) - 0| = |x| < \epsilon.$$ 2. **For rational points** $$(x \in \mathbb{Q})$$: If $$x$$ is rational, then $$F(x) = 0$$, so: $$|F(x) - 0| = 0 < \epsilon,$$ which always holds. Since every neighborhood around $$x = 0$$ contains both rational and irrational points, and the function value is always within $$\epsilon$$ of $$0$$, the limit exists and equals $$0$$. $$\lim_{x \to 0} F(x) = 0.$$ --- ### **(b) Show that if $$a \neq 0$$, then** $$\lim_{x \to a} F(x)$$ **does not exist.** **Proof:** Suppose $$a \neq 0$$. To show that the limit does not exist, we must demonstrate that the function values $$F(x)$$ do not approach a single value as $$x \to a$$. 1. **Consider rational points near $$a$$**: If $$x \in \mathbb{Q}$$, then $$F(x) = 0$$. 2. **Consider irrational points near $$a$$**: If $$x \notin \mathbb{Q}$$, then $$F(x) = x$$. As $$x \to a$$, these values approach $$a$$. 3. **Contradiction in approaching a limit**: - For any sequence of rational numbers $$\{x_n\} \to a$$, we have $$F(x_n) = 0$$. - For any sequence of irrational numbers $$\{y_n\} \to a$$, we have $$F(y_n) = y_n \to a$$. Since these two subsequences approach different values ($$0$$ and $$a$$, respectively), the function does not settle to a single value as $$x \to a$$. Therefore, the limit: $$\lim_{x \to a} F(x) \text{ does not exist if } a \neq 0.$$ --- ### Do you want further clarification or details? --- ### Related Questions: 1. What is the difference between pointwise and uniform continuity for $$F(x)$$? 2. How does the density of rationals and irrationals affect the behavior of $$F(x)$$? 3. Can the modified Dirichlet function be integrated on an interval $$[0, 1]$$? 4. Is $$F(x)$$ continuous at any point? Why or why not? 5. What is the nature of the discontinuities of $$F(x)$$? --- ### **Tip:** When evaluating limits of piecewise functions, analyze the behavior along different subsequences to check for contradictions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Real Analysis
Limits
Piecewise Functions
Rational and Irrational Numbers
Dirichlet Function
Formulas
$F(x) = x$ for $x \notin \mathbb{Q}$
$F(x) = 0$ for $x \in \mathbb{Q}$
$\lim_{x \to a} F(x)$
Theorems
Limit of Piecewise Functions
Density of Rational and Irrational Numbers
Suitable Grade Level
Undergraduate
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