Piecewise question, Please Help

[gwslawndart]

Hey guys, Would really appreciate your help with this answer, I am not quite sure how to go about answering this. Thanks in advance


Let f(x)=----{ x if x is a rational number
--------------{x² if x is an irrational number

lim f(x)
x->a

For what values of a, if any, does lim f(x) exist? Be sure to justify your response.

Luis

 

[pka]

Try to convince yourself that there two values: a=0 & a=1.

 

[gwslawndart]

"Try to convince yourself that there two values: a=0 &a

After giving this problem some thought, i graphed both x and x^2, i graphed X^2 as a dotted curve to represent only irrational values, and X as a line to represent only rational numbers. (is this correct?) I did this because, this function however not precisely equal to X=0 and X=1 because the values x=0 and x=1 are both rational numbers, however the graph of this piecewise function does approach infinitely close in 2 location, x=0 and x=1, because the portion of the graph which corresponds to x'2 is every irrational number and the graph of x is every rational number on the real number line. The lim exists at only these 2 values of X, because what we know of irrational and rational number is that, there are inifinitely many irrational numbers between rational numbers, and in this case the numbers as X--A come arbitraliy close to X=0 and X=1.

Is this the right answer? I feel like i was being repitative, but the request from my professor was to explain it thoroughly.

Again, thanks for your answers and help

Luis

 

[pka]

\(\displaystyle \L x \approx 0\quad \Rightarrow \quad x^2 \approx 0\quad \& \quad x \approx 1\quad \Rightarrow \quad x^2 \approx 1\)
Those are the only two values that have that property.

 

[Deleted member 4993]

pka,

Isn't it true for this function that the function is discontinuous everywhere - but limit exists everywhere?

The function has a definite value at every point. Thus I would think limit exists.

It is discontinuous everywhere because we could not define a interval for epsilon-delta test.

Am I thinking correctly about this problem?

 

[pka]

It is discontinuous everywhere because we could not define a interval for epsilon-delta test.

Here is a proof that it is continuous at x=1.
If \(\displaystyle \varepsilon > 0, \Rightarrow \quad \delta = \min \left\{ {1,\frac{\varepsilon }{3}} \right\}\) and \(\displaystyle \left| {x - 1} \right| < \delta \Rightarrow \quad \left| {x + 1} \right| < 3\).
Now we have \(\displaystyle \left| {f(x) - 1} \right| = \left\{ {\begin{array}{lc}
{\left| {x - 1} \right|} & {x \in Q} \\
{\left| {x^2 - 1} \right|} & {x \notin Q} \\
\end{array}} \right.\)

In the first case we have \(\displaystyle \left| {x - 1} \right| < \delta = \varepsilon \Rightarrow
\quad \left| {f(x) - 1} \right| < \varepsilon\).

In the second case we have \(\displaystyle \left| {x - 1} \right| < \delta \le 1 \Rightarrow \quad \left| {x^2 - 1} \right| = \left| {x + 1} \right|\left| {x - 1} \right| < 3(\delta ) < \varepsilon\).

QED

 

[Deleted member 4993]

Thanks, pka.

So the function is continuous -only - at the vicinity of x = 0 and x = 1.(I could not find that before).

However, the function has defined limit at every point though - correct?

 

[pka]

So the function is continuous only AT = 0 and x = 1.(I could not find that before).

Yes that is correct.

However, the function has defined limit at every point though - correct?

No that is incorrect.

Let’s take an example, x=2. The sequence \(\displaystyle \left( {2 + \frac{{\sqrt 2 }}{n}} \right) \to 2\).
But every term is irrational so \(\displaystyle f\left( {2 + \frac{{\sqrt 2 }}{n}} \right) = \left( {2 + \frac{{\sqrt 2 }}{n}} \right)^2 = \left( {4 + \frac{{2\sqrt 2 }}{n} + \frac{2}{{n^2 }}} \right) \to 4\).

On the other hand, the sequence \(\displaystyle \left( {2 + \frac{1}{n}} \right) \to 2\) and every term is rational.
Thus, \(\displaystyle f\left( {2 + \frac{1}{n}} \right) = \left( {2 + \frac{1}{n}} \right) \to 2\).

In order for f to have a limit at 2, say L, for any number close to 2, but not 2, say t, then f(t) must be close to L. But that is false.

Because any real number is the limit of a sequence of rationals and also the limit of irrationals then the only time the limit will exist is when x=0 or x=1.
Because this requires \(\displaystyle x = x^2\).

 

[gwslawndart]

So, Was my answer correct? =/?

I appreciate the input from everyone, however, i am not sure whether my previous explaination of the answer, is correct or incorrect,

Thanks,

Luis

 

[pka]

Re: So, Was my answer correct? =/?

however, i am not sure whether my previous explanation of the answer, is correct or incorrect

Frankly I do not know if it is correct or not.
I cannot read it. Or that is, I do not understand what in the world you are trying to say. If I were you, I would try to make a more mathematical statement.