Limit of piecewise function using epsilon delta

[songoku]

Homework Statement

Prove that f(x) is continuous at x = 2 where
$f(x)=\begin{cases} x^3 +2, & x \lt 2 \\ 10, & x =2 \\ x^2+6, & x > 2 \end{cases}$

Relevant Equations

$\lim_{x \to a^{-}} f(x)=L$ if for every $\epsilon>0$ there is $\delta >0$ such that if $a-\delta<x<a$ then $|f(x)-L|<\epsilon$

$\lim_{x \to a^{+}} f(x)=L$ if for every $\epsilon>0$ there is $\delta >0$ such that if $a<x<a+\delta$ then $|f(x)-L|<\epsilon$

1) proofing $\lim_{x \to 2^{-} f(x)=10}$

$$|x^3 +2-10|<\epsilon$$
$$-\epsilon<x^3-8<\epsilon$$
$$8-\epsilon<x^3<8+\epsilon$$
$$\sqrt[3] {8-\epsilon}<x<\sqrt[3] {8+\epsilon}$$
$$\sqrt[3] {8-\epsilon}-2<x-2<\sqrt[3] {8+\epsilon}-2$$
$$|\sqrt[3] {8-\epsilon}-2|<|x-2|<|\sqrt[3] {8+\epsilon}-2|$$

I want to take the minimum of ${|\sqrt[3] {8-\epsilon}-2|, |\sqrt[3] {8+\epsilon}-2|}$ as $\delta$ but how to know which one without using calculator?


2) proofing $\lim_{x \to 2^{+} f(x)=10}$

$$|x^2+6-10|<\epsilon$$
$$-\epsilon<x^2-4<\epsilon$$
$$4-\epsilon<x^2<4+\epsilon$$
$$\sqrt{4-\epsilon}<x<\sqrt{4+\epsilon}$$
$$\sqrt{4-\epsilon}-2<x-2<\sqrt{4+\epsilon}-2$$
$$|\sqrt{4-\epsilon}-2|<|x-2|<|\sqrt{4+\epsilon}-2|$$

Same thing, how to know the minimum of ${|\sqrt{4-\epsilon}-2| , |\sqrt{4+\epsilon}-2|}$ without calculator?

Thanks

 

[pasmith]

You haven't told us the definition of $f$.

You can assume $\epsilon < 8$ (a $\delta$ which works for such an $\epsilon$ will also work for any larger $\epsilon$) so that $0 < 8 - \epsilon < 8 + \epsilon$.

 

[songoku]

You haven't told us the definition of $f$.

Sorry I don't understand. Isn't $f$ defined in the question?

For $x\to 2^{-}, f(x) = x^3 +2$ and for $x\to 2^{+}, f(x)=x^2 + 6$. Is this what you mean?

You can assume $\epsilon < 8$ (a $\delta$ which works for such an $\epsilon$ will also work for any larger $\epsilon$) so that $0 < 8 - \epsilon < 8 + \epsilon$.

Ok so the the right side limit, I also assume $0<\epsilon<4$ so the minimum would be $|\sqrt{4+\epsilon}-2|$

Is it ok to have two $\delta$, one for each one-sided limit?

Thanks

 

[PeroK]

You can do the two one-sided limits as separate calculations. The limit is $L$ if and only if both one-sided limits are $L$.

 

[songoku]

You can do the two one-sided limits as separate calculations. The limit is $L$ if and only if both one-sided limits are $L$.

How to continue the working to complete the proof?

For left side limit, let $\delta=|\sqrt[3] {8+\epsilon}-2|$. Taking $\delta>0$ and $0<\epsilon<8$ , then $\delta=|\sqrt[3] {8+\epsilon}-2|=\sqrt[3] {8+\epsilon}-2$

So:
$$2-(\sqrt[3] {8+\epsilon}-2)<x<2$$
$$4-\sqrt[3] {8+\epsilon}<x<2$$

How to change this into $|x^3 -8|<\epsilon$?

Thanks

 

[PeroK]

You need to establish $\delta$ for each $\epsilon$. You should stay focused on $|x -2|$.

 

[PeroK]

Alternatively, go back to the requirement that:
$$|x^2 - 4| < \epsilon$$Now factorise the left hand side. That's the normal approach.

 

[songoku]

You need to establish $\delta$ for each $\epsilon$

For left side limit $(x\to 2^{-})$, the $\delta=|\sqrt[3] {8+\epsilon}-2|$

For right side limit, $(x\to 2^{+})$, the $\delta=|\sqrt{4+\epsilon}-2|$

Is this what you mean?

You should stay focused on $|x -2|$.

Yeah I am having difficulty getting the $|x-2|$ to complete the proof. I am trying to use $a-\delta<x<a$ for left side limit and $a<x<a+\delta$ for right side limit

Alternatively, go back to the requirement that:
$$|x^2 - 4| < \epsilon$$Now factorise the left hand side. That's the normal approach.

You mean for right side limit?

$$|x^2 -4|<\epsilon$$
$$|x-2||x+2|<\epsilon$$
$$|x-2|<\frac{\epsilon}{|x+2|}$$

Let: $\delta=1$ so $|x-2|<1 \to 3<|x+2|<5$, then:

$$|x-2|<\frac{\epsilon}{|x+2|}$$
$$|x-2|<\frac{\epsilon}{5}$$

So take $\delta=\text{min}~(1, \frac{\epsilon}{5})$

$$|x^2 - 4|=|x-2||x+2|<\frac{\epsilon}{5}.5=\epsilon$$

Is this what you mean?

Thanks

 

[FactChecker]

I want to take the minimum of ${|\sqrt[3] {8-\epsilon}-2|, |\sqrt[3] {8+\epsilon}-2|}$ as $\delta$ but how to know which one without using calculator?

Good work. You don't need to know what it is, as long as you know it is positive and small enough. Just set $\delta_1 = \min( {|\sqrt[3] {8-\epsilon}-2|, |\sqrt[3] {8+\epsilon}-2|})$.

Same thing, how to know the minimum of ${|\sqrt{4-\epsilon}-2| , |\sqrt{4+\epsilon}-2|}$ without calculator?

Same answer. Set $\delta_2 = \min({|\sqrt[3] {8-\epsilon}-2|, |\sqrt[3] {8+\epsilon}-2|})$.
Now set $\delta = \min( \delta_1, \delta_2)$ to show continuity from both sides.

 

[songoku]

Good work. You don't need to know what it is, as long as you know it is positive and small enough. Just set $\delta_1 = \min( {|\sqrt[3] {8-\epsilon}-2|, |\sqrt[3] {8+\epsilon}-2|})$.

Same answer. Set $\delta_2 = \min({|\sqrt[3] {8-\epsilon}-2|, |\sqrt[3] {8+\epsilon}-2|})$.

Wait, I don't need to choose which one is the minimum? But then when I want to prove it, what should I put as $delta$?

Now set $\delta = \min( \delta_1, \delta_2)$ to show continuity from both sides.

Oh I have to take one common $\delta$ and use it for left and right side limit?

Thanks

 

[PeroK]

Wait, I don't need to choose which one is the minimum? But then when I want to prove it, what should I put as $delta$?

In this case, you might be able to calculate which one is the minimum. In general, however, you just need to show that such a delta exists. It's common to have something like:

Let $\delta = min \{1, \delta_1, \delta_2\}$.

 

[BvU]

In this case, you might be able to calculate which one is the minimum. In general, however, you just need to show that such a delta exists. It's common to have something like:

Let δ=min{1,δ1,δ2}.

 

[FactChecker]

Wait, I don't need to choose which one is the minimum? But then when I want to prove it, what should I put as $delta$?

No. You only need to prove that there exists a positive delta that satisfies the requirements. You don't need to say anything more about it. You can leave delta expressed as a minimum between two (or more) strictly positive numbers. You know that a minimum exists and you don't need to say which one is the minimum.

 

[songoku]

In this case, you might be able to calculate which one is the minimum. In general, however, you just need to show that such a delta exists. It's common to have something like:

Let $\delta = min \{1, \delta_1, \delta_2\}$.

No. You only need to prove that there exists a positive delta that satisfies the requirements. You don't need to say anything more about it. You can leave delta expressed as a minimum between two (or more) positive numbers. You know that a minimum exists and you don't need to say which one is the minimum.

Then how to continue the proof?

Let:
$\delta_1=\text{min}\{\sqrt[3] {8-\epsilon}-2, \sqrt[3] {8+\epsilon}-2\}$
$\delta_2=\text{min}\{\sqrt{4-\epsilon}-2, \sqrt {4+\epsilon}-2\}$

$\delta=\text{min} \{\delta_1, \delta_2 \}$

For left limit: $2-\delta<x<2$. How to change this into $|x^3 - 8|<\epsilon?$

Thanks

 

[Gavran]

I have used an online graphing calculator tool and I have found that $$ \delta=|\sqrt[3]{8+\epsilon}-2| $$. I am doubtful about finding $\delta$ easily. By the way, the original poster can always follow the recommendation in the post #13.

 

[songoku]

By the way, the original poster can always follow the recommendation in the post #13.

I am trying to but sorry I don't have idea how to continue without choosing a certain value for $\delta$. What I practice up until now is the one which always choose the value of $\delta$.

If not choosing any value for $\delta$, what is the continuation after $\delta=\text{min} \{\delta_1, \delta_2\}$?

Thanks

 

[PeroK]

Then how to continue the proof?

Let:
$\delta_1=\text{min}\{\sqrt[3] {8-\epsilon}-2, \sqrt[3] {8+\epsilon}-2\}$
$\delta_2=\text{min}\{\sqrt{4-\epsilon}-2, \sqrt {4+\epsilon}-2\}$

$\delta=\text{min} \{\delta_1, \delta_2 \}$

For left limit: $2-\delta<x<2$. How to change this into $|x^3 - 8|<\epsilon?$

Thanks

The first number in each set is negative!

 

[FactChecker]

Then how to continue the proof?

You are done. You have proved that $\delta \gt 0$ exists which satisfies the requirement.

Let:
$\delta_1=\text{min}\{\sqrt[3] {8-\epsilon}-2, \sqrt[3] {8+\epsilon}-2\}$
$\delta_2=\text{min}\{\sqrt{4-\epsilon}-2, \sqrt {4+\epsilon}-2\}$

You accidentally dropped the absolute values.

$\delta=\text{min} \{\delta_1, \delta_2 \}$

Right.

For left limit: $2-\delta<x<2$. How to change this into $|x^3 - 8|<\epsilon?$

This is unnecessary for the proof.

 

[PeroK]

You are done. You have proved that $\delta \gt 0$ exists which satisfies the requirement.

You accidentally dropped the absolute values.

Right.

This is unnecessary for the proof.

Although, there should only be one $\delta$ in each set, as we are dealing with a one-sided limit. For example, for the righthand limit we have:
$$2<x < \sqrt{4+\epsilon}$$The other inequality is superfluous.

 

[FactChecker]

Although, there should only be one $\delta$ in each set, as we are dealing with a one-sided limit. For example, for the righthand limit we have:
$$2<x < \sqrt{4+\epsilon}$$The other inequality is superfluous.

For two-sided continuity, the limits must be equal. On each side, that delta and any smaller, strictly positive, delta will do. So the minimum works for both sides simultaneously for the same limit value.

The OP was trying to get more specific about exact delta values than the proof needed.

 

[PeroK]

For two-sided continuity, the limits must be equal. On each side, that delta and any smaller, strictly positive, delta will do. So the minimum works for both sides simultaneously for the same limit value.

The OP was trying to get more specific about exact delta values than the proof needed.

If you look at his calculations. For $x > 2$, where the function is quadratic, he includes the case $x < 2$. Similarly, when $x <2$, he includes a calculation for $x >2$ for the cubic.

 

[songoku]

The first number in each set is negative!

You accidentally dropped the absolute values.

Ah my bad

You are done. You have proved that $\delta \gt 0$ exists which satisfies the requirement.

This is unnecessary for the proof.

Oh ok, but how about something like this?

That example is from textbook and there is part 2 of the working (to show that this $\delta$ works). Is this part not needed and I can just stop at part 1 after stating $\delta=\frac{\epsilon}{4}$?

If you look at his calculations. For $x > 2$, where the function is quadratic, he includes the case $x < 2$. Similarly, when $x <2$, he includes a calculation for $x >2$ for the cubic.

I am sorry I don't get this. In post #1, when dealing with left side limit, I use the cubic one and not using the quadratic at all. When dealing with right side limit, I use the quadratic one, not using the cubic at all.

Do I misinterpret your post?

Thanks

 

[PeroK]

You calculated the bounds based on a two-sided limit. Even though you said you were doing a one-sided limit. That's why you ended up with two sets of two deltas.

 

[Gavran]

$$\sqrt[3] {8-\epsilon}-2<x-2<\sqrt[3] {8+\epsilon}-2$$

$x-2\lt0$ and $\sqrt[3]{8+\epsilon}-2\gt0$ implies that $x-2\lt\sqrt[3]{8+\epsilon}-2$ always holds so $\delta_1=2-\sqrt[3]{8-\epsilon}$.

$$\sqrt{4-\epsilon}-2<x-2<\sqrt{4+\epsilon}-2$$

For $\epsilon\le4$
$x-2\gt0$ and $\sqrt{4-\epsilon}-2\lt0$ implies that $x-2\gt\sqrt{4-\epsilon}-2$ always holds so $\delta_2=\sqrt{4+\epsilon}-2$.
For $\epsilon\gt4$
$\delta_2=\sqrt{4+\epsilon}-2$ is the only option because $(\sqrt{4-\epsilon}-2)\notin\mathbb{R}$.

 

[FactChecker]

Oh ok, but how about something like this?

View attachment 362800
That example is from textbook and there is part 2 of the working (to show that this $\delta$ works). Is this part not needed and I can just stop at part 1 after stating $\delta=\frac{\epsilon}{4}$?

I can't say with authority whether part 2 is required. I am not a "proof policeman". If part 1 had stated things differently, like using the minimum rather than saying $\delta = \epsilon/4$ was a "guess", I would have considered part 1 to be enough of a proof.

 

[songoku]

You calculated the bounds based on a two-sided limit. Even though you said you were doing a one-sided limit. That's why you ended up with two sets of two deltas.

Is calculating the bounds based on one - sided limit like post #24?

$x-2\lt0$ and $\sqrt[3]{8+\epsilon}-2\gt0$ implies that $x-2\lt\sqrt[3]{8+\epsilon}-2$ always holds so $\delta_1=2-\sqrt[3]{8-\epsilon}$.

But in post#15, the delta you take is $|\sqrt[3]{8+\epsilon}-2|$, not the negative one?

I can't say with authority whether part 2 is required. I am not a "proof policeman". If part 1 had stated things differently, like using the minimum rather than saying $\delta = \epsilon/4$ was a "guess", I would have considered part 1 to be enough of a proof.

By minimum, do you mean $\delta=\text{min}~(1, \frac{\epsilon}{4})$?

Thanks

 

[anuttarasammyak]

Please find below the plots of my understanding the situation. I hope it is not wrong and has some meanings.

 

[PeroK]

In general, for a right hand limit at $a$ you are looking for:$$a < x <a +\delta$$And not $$0<|x-a|< \delta$$PS I see this was in the OP. The difficulties arose from not simply taking the one-sided limits separately.

 

[Gavran]

But in post#15, the delta you take is |8+ϵ3−2|, not the negative one?

The post #15 was my mistake.

When $x\to 2^-$
$\sqrt[3]{8-\epsilon}-2\lt x-2\lt\sqrt[3]{8+\epsilon}-2$ does not imply $|\sqrt[3]{8-\epsilon}-2|\lt |x-2|\lt|\sqrt[3]{8+\epsilon}-2|$
and when $x\to 2^+$
$\sqrt{4-\epsilon}-2\lt x-2\lt\sqrt{4+\epsilon}-2$ does not imply $|\sqrt{4-\epsilon}-2|\lt |x-2|\lt|\sqrt{4+\epsilon}-2|$.

When $x\to2^-$ you have
$\sqrt[3]{8+\epsilon}-2\gt0$,
$\sqrt[3]{8-\epsilon}-2\lt0$
and
$x-2\lt0$,
$x-2\lt\sqrt[3]{8+\epsilon}-2$,
$x-2\gt\sqrt[3]{8-\epsilon}-2$.
These five inequalities imply $\sqrt[3]{8-\epsilon}-2\lt x-2\lt0$.

Also, when $x\to2^+$ and $\epsilon\le4$ you have
$\sqrt{4+\epsilon}-2\gt0$,
$\sqrt{4-\epsilon}-2\lt0$
and
$x-2\gt0$,
$x-2\lt\sqrt{4+\epsilon}-2$,
$x-2\gt\sqrt{4-\epsilon}-2$.
These five inequalities imply $0\lt x-2\lt\sqrt{4+\epsilon}-2$.
For $x\to2^+$ and $\epsilon\gt4$
$\sqrt{4-\epsilon}-2\lt0$ and $x-2\gt\sqrt{4-\epsilon}-2$ are not defined, so you have
$\sqrt{4+\epsilon}-2\gt0$
and
$x-2\gt0$,
$x-2\lt\sqrt{4+\epsilon}-2$.
These three inequalities imply $0\lt x-2\lt\sqrt{4+\epsilon}-2$.

Now, for $x\to2^-$ there is $0\lt|x-2|\lt2-\sqrt[3]{8-\epsilon}$ and for $x\to2^+$ there is $0\lt|x-2|\lt\sqrt{4+\epsilon}-2$.

 

[PeroK]

That's not really how we are supposed to construct a proof here. Instead, for $0<\epsilon <1$, we should let:
$$\delta = min\{\delta_1, \delta_2 \}$$Where
$$\delta_1 = 2 - \sqrt[3]{8-\epsilon}, \ \delta_2 = \sqrt{4+\epsilon} -2$$Then:$$0<|x -2|<\delta$$$$ \implies 2 - \delta < x <2 \ \text{or} \ 2 < x < 2+ \delta$$$$\implies \sqrt[3]{8-\epsilon} < x <2 \ \text{or} \ 2 < x < \sqrt{4+\epsilon}$$etc.