Piecewise function - rational and irrational

[Dethrone]

$$g(x)=\begin{cases}x^2, & \text{ if x is rational} \\[3pt] 0, & \text{ if x is irrational} \\ \end{cases}$$

a) Prove that $\lim_{{x}\to{0}}g(x)=0$
b) Prove also that $\lim_{{x}\to{1}}g(x) \text{ D.N.E}$

I've never seen a piecewise function defined that way...hints?

 

[magneto1]

Hint: Use the $\delta-\varepsilon$ definition of limit.

Let $\varepsilon > 0$ be given, choose $\delta = [...]$. Consider the cases of $\left| f(x) - f(0) \right|$ if $x$ is rational and if $x$ is irrational.

 

[Dethrone]

Darn! (Angry) I was planning to skip all the $\delta-\varepsilon$ questions because I haven't learned it yet, but seeing as you put in the time to answer my question and that I've already asked it, I will learn it first thing tomorrow morning (Nod)

 

[mathbalarka]

Here's a restate of the $\epsilon-\delta$ proof in a little nonrigorous language, see if you can verify it step-by-step :

Let $g_0$ be the function $g$ restricted to the set of rationals $\Bbb Q$. As $g_0(x)$ is just $x^2$ and $\lim_{x \to 0} x^2 = 0$, the limit point of $\{g_0(x_n)\}$ is $0$ for any sequence of rationals $\{x_n\}$ converging to $0$.

On the other hand let $g_1$ be the zero-map $x \mapsto 0$ restricted to the set of irrationals $\Bbb R - \Bbb Q$. As $\lim_{x \to 0} 0 = 0$, the limit point of $\{f(y_n)\}$ is $0$ for any sequence of irrationals $\{y_n\}$ converging to $0$.

Thus, as $g$ is the aggregate of the two functions $g_0$ and $g_1$, $\{g(z_n)\}$ has limit point $0$ for any sequence $\{z_n\}$ converging to $0$. A proper continuous choice of $\{z_n\}$ verifies that $\lim_{x \to 0} g(x)$ does indeed converge to $0$.

 

[Euge]

$$g(x)=\begin{cases}x^2, & \text{ if x is rational} \\[3pt] 0, & \text{ if x is irrational} \\ \end{cases}$$

a) Prove that $\lim_{{x}\to{0}}g(x)=0$
b) Prove also that $\lim_{{x}\to{1}}g(x) \text{ D.N.E}$

I've never seen a piecewise function defined that way...hints?

Hi Rido12,

For part a), note that $0\le g(x) \le x^2$ for all $x\in \Bbb R$. So by the squeeze theorem, $lim_{x\to 0} g(x) = 0$.

For part b), you can argue by contradiction, using the fact that every interval contains an irrational number. Suppose $\lim_{x \to 0} g(x)$ exists with limit $L$. Then for every $\epsilon > 0$, there exists a $\delta > 0$ such that for all $x$, $|x - 1| < \delta$ implies $|g(x) - L| < \epsilon$. Choose $\epsilon < 1/2$. If $L = 0$, then taking $x = 1$ forces $|g(x) - L| = 1 >\epsilon$. So $L$ is nonzero. Take an irrational number $x$ in $(1 - \delta, 1 + \delta)$. Then $|g(x) - L| < \epsilon$, whhich implies $|L| < \epsilon$. On the other hand, $|g(1) - L| < \epsilon$, that is, $|1 - L| < \epsilon$. Hence $\epsilon > |L| > 1 - \epsilon$, which yields $\epsilon > 1/2$ $(\rightarrow \leftarrow)$.