1. ## Prove discontinuity at

Define $\displaystyle f:\mathbb{R}\to\mathbb{R}$ by $\displaystyle f(x)=8x, \ \text{if} \ x\in\mathbb{Q}$ and $\displaystyle f(x)=2x^2+8$ if x is irrational. Prove from the definition that f is discontinuous at 1.

Let $\displaystyle \epsilon >0$. Then there exist $\displaystyle \delta_1, \ \delta_2>0$.

$\displaystyle |8x-8|=8|x-1|<\epsilon\Rightarrow |x-1|<\frac{\epsilon}{8}$.

Let $\displaystyle \delta_1=\frac{\epsilon}{8}$.

$\displaystyle |2x^2+8-10|=2|x^2-6|<\epsilon$.

Does it suffice to say that $\displaystyle |x^2-6|$ can't be manipulated to some form of $\displaystyle |x-1|$ now?

2. ## Re: Prove discontinuity at

Originally Posted by dwsmith
Does it suffice to say that $\displaystyle |x^2-6|$ can't be manipulated to some form of $\displaystyle |x-1|$ now?
No, it is not. Take into account that $\displaystyle |f(x)-f(1)|=\begin{Bmatrix}8|x-1| & \mbox{ if }& x\in \mathbb{Q}\\2x^2 & \mbox{if}& x\not\in \mathbb{Q}\end{matrix}$

Now, choose for instance $\displaystyle \epsilon =1$ . If $\displaystyle |x-1|<\epsilon/8$ then $\displaystyle 2x^2>\epsilon$ .

3. ## Re: Prove discontinuity at

Originally Posted by dwsmith
Define $\displaystyle f:\mathbb{R}\to\mathbb{R}$ by $\displaystyle f(x)=8x, \ \text{if} \ x\in\mathbb{Q}$ and $\displaystyle f(x)=2x^2+8$ if x is irrational. Prove from the definition that f is discontinuous at 1.

Let $\displaystyle \epsilon >0$. Then there exist $\displaystyle \delta_1, \ \delta_2>0$.

$\displaystyle |8x-8|=8|x-1|<\epsilon\Rightarrow |x-1|<\frac{\epsilon}{8}$.

Let $\displaystyle \delta_1=\frac{\epsilon}{8}$.

$\displaystyle |2x^2+8-10|=2|x^2-6|<\epsilon$.

Does it suffice to say that $\displaystyle |x^2-6|$ can't be manipulated to some form of $\displaystyle |x-1|$ now?
I don't know how formal they want you to be but you could just say say that since there is a sequence of rationals and a sequence of irrationals converging to any point and continuous functions preserve sequences that any point $latex x_0$ where $latex f$ is continuous would have to satisfy $latex 8x_0=2x_0^2+8$--and $latex 1$ does not.

4. ## Re: Prove discontinuity at

Originally Posted by Drexel28
I don't know how formal they want you to be but you could just say say that since there is a sequence of rationals and a sequence of irrationals converging to any point and continuous functions preserve sequences that any point $latex x_0$ where $latex f$ is continuous would have to satisfy $latex 8x_0=2x_0^2+8$--and $latex 1$ does not.
I suppose they don't want to use that characterization because the problem says "Prove from the definition".

5. ## Re: Prove discontinuity at

Originally Posted by dwsmith
Define $\displaystyle f:\mathbb{R}\to\mathbb{R}$ by $\displaystyle f(x)=8x, \ \text{if} \ x\in\mathbb{Q}$ and $\displaystyle f(x)=2x^2+8$ if x is irrational. Prove from the definition that f is discontinuous at 1.

Let $\displaystyle \epsilon >0$. Then there exist $\displaystyle \delta_1, \ \delta_2>0$.

$\displaystyle |8x-8|=8|x-1|<\epsilon\Rightarrow |x-1|<\frac{\epsilon}{8}$.

Let $\displaystyle \delta_1=\frac{\epsilon}{8}$.

$\displaystyle |2x^2+8-10|=2|x^2-6|<\epsilon$.

Does it suffice to say that $\displaystyle |x^2-6|$ can't be manipulated to some form of $\displaystyle |x-1|$ now?
Dear dwsmith,

Using the Epsilon Delta definition of continuity, we can show that this function is discontinuous at x=1.

Epsilon-delta definition of continuity:

$\displaystyle {f(x)~:~D\rightarrow\Re\mbox{ is continuous at x=c iff }\forall~\epsilon>0~\exists~\delta>0\mbox{ such that }|f(x)-f(c)|<\epsilon \mbox{ whenever }|x-c|<\delta}$.

In this particular problem, c=1.

$\displaystyle f(x)=\left\{\begin{array}{cll}8x & \mbox{if}&x\in Q\\2x^{2}+8 & \mbox{if} & x\not\in Q \end{array}\right.$

$\displaystyle |f(x)-f(1)|=\begin{Bmatrix}8|x-1| & \mbox{ if }& x\in Q\\2x^2 & \mbox{if}& x\not\in Q\end{matrix}$

Let $\displaystyle \epsilon=0.08$ and we shall try to find a $\displaystyle \delta$

Therefore, $\displaystyle 8|x-1|<0.08\Rightarrow |x-1|<0.01\mbox{ when }x\in Q$

$\displaystyle 2x^{2}<0.08\Rightarrow 0.8<|x-1|<1.2\mbox{ when }x\not\in Q$

We should find a $\displaystyle \delta$ that satisfies for all the x values may they be rational or irrational. But this is impossible as the two intervals $\displaystyle |x-1|<0.01\mbox{ and }0.8<|x-1|<1.2$ does not have a intersection.

That is for $\displaystyle \epsilon=0.08$ a $\displaystyle \delta$ value does not exist. Therefore the function is discontinuous at x=1.

Hope you understood my reasoning. And may there be any doubt please don't hesitate to ask.

6. ## Re: Prove discontinuity at

the general strategy to prove f is NOT continuous at some point a, is to find SOME ε > 0, for which we can't find ANY δ > 0,

for which |x - a| < δ implies |f(x) = f(a)| < ε. now let's think about this:

8x and 2x^2 + 8 are both continuous functions. the first is "near" 8 for x "near" 1, and the second is "near" 10, for x "near 1".

this suggests we might be able to use some 0 < ε < 2. let's try it for ε = 1.

in order to know that 8x is within 1 of 8, we need 7 < 8x < 9, that is 7/8 < x < 9/8. that is: if |x - 1| < 1/8,

then |8x - 8| < 1. that's pretty straight-forward. it should be clear that allowing δ > 1/8 won't work,

then for some of the (rational) x's in (1-δ, 1+δ), f(x) will be too far away from f(1) = 8.

now |x - 1| < 1/8 is just saying 7/8 < x < 9/8. so 49/64 < x^2 < 81/64, so 98/64 < 2x^2 < 162/64,

so 610/64 < 2x^2 + 8 < 674/64. but 9 < 610/64, so for all x with |x - 1| < 1/8,

for all of the rational x's, f(x) is within 1 of f(1), and this is true for NONE of the irrational x's with |x - 1| < 1/8.

so no 0 < δ ≤ 1/8 will work (for ε = 1), and no δ > 1/8 will work, either (since SOME x's in (1-δ, 1+δ) will be

make f(x) too far away). so, there's one ε > 0, for which it is impossible to find a suitable δ.