Use the Squeeze Theorem

[Yoyoma2732]

Please help solve this problem.

 

[Dr.Peterson]

Please help solve this problem.
View attachment 28760

What help do you need? What have you tried?

I'd look for a function that both [imath]1+2x^2[/imath] and [imath]1+x^4[/imath] are less than, for all x in (-1,1). (A function that they are both greater than is easy.)

 

[Yoyoma2732]

Thank you Dr. Peterson,
I had difficultly deciphering what “x is rational” and “x is irrational” meant and finding a place to start from there. From your advise, this is my attempt at the problem. Please let me know if my reasoning is flawed. Thank you for your help thus far.

What help do you need? What have you tried?

I'd look for a function that both [imath]1+2x^2[/imath] and [imath]1+x^4[/imath] are less than, for all x in (-1,1). (A function that they are both greater than is easy.)

 

[Dr.Peterson]

Thank you Dr. Peterson,
I had difficultly deciphering what “x is rational” and “x is irrational” meant and finding a place to start from there. From your advise, this is my attempt at the problem. Please let me know if my reasoning is flawed. Thank you for your help thus far.

None of your work is irrelevant. In a piecewise function, the two "pieces" can be utterly unrelated, and they never apply to the same value of x. So subtracting one formula from the other is meaningless.

Furthermore, your "bounds" are not on the function f, and the limit you state doesn't answer the question, which is about the limit of f.

It sounds like you don't understand how the function works. You might think about it by lightly graphing [imath]y = 1+2x^2[/imath] and [imath]y=1+x^4[/imath]. Then the idea of the function in principle is that for every rational number value of x, you would put a dot on the first curve, and for every other value of x you would put a dot on the second curve. Of course, both kinds of numbers are scattered densely along the number line, so there's no way to actually do that; the theoretical graph would look essentially like what you've drawn, just two lightly drawn curves!

So you can tell from the graph that the limit is 1; and what you see is more or less the way to prove it. Which curve is higher, for x near zero? You can use that as an upper bound, since the value of f(x) is never greater than that. Which curve is lower? You can use that as a lower bound. Now the limit of f(x) is between the limits of those two functions! Give it a squeeze, and you have your proof!

 

[Steven G]

1.5 is rational, so f(1.5) = 1 + 2(1.5^2) = 1 + 2(2.25) = 1 + 4.5= 5.5. Notice that I used the rule 1 + 2x^2 instead of 1+x^4 since the x that I used was rational

1/sqrt(2) is irrational, so f(1/sqrt(2)) = 1 + (1/sqrt(2))^4= 1 + 1/4 = 5/4. Note that I used the rule 1+x^4 to find f(1/sqrt(2)) since 1/sqrt(2) is irrational.

 

[Yoyoma2732]

None of your work is irrelevant. In a piecewise function, the two "pieces" can be utterly unrelated, and they never apply to the same value of x. So subtracting one formula from the other is meaningless.

Furthermore, your "bounds" are not on the function f, and the limit you state doesn't answer the question, which is about the limit of f.

It sounds like you don't understand how the function works. You might think about it by lightly graphing [imath]y = 1+2x^2[/imath] and [imath]y=1+x^4[/imath]. Then the idea of the function in principle is that for every rational number value of x, you would put a dot on the first curve, and for every other value of x you would put a dot on the second curve. Of course, both kinds of numbers are scattered densely along the number line, so there's no way to actually do that; the theoretical graph would look essentially like what you've drawn, just two lightly drawn curves!

So you can tell from the graph that the limit is 1; and what you see is more or less the way to prove it. Which curve is higher, for x near zero? You can use that as an upper bound, since the value of f(x) is never greater than that. Which curve is lower? You can use that as a lower bound. Now the limit of f(x) is between the limits of those two functions! Give it a squeeze, and you have your proof!

Thank you for your help! Based upon what you said, this is what I came up with. Again, please let me know if I applied what you were trying to explain to me.

 

[Yoyoma2732]

Thank you for your help! Based upon what you said, this is what I came up with. Again, please let me know if I applied what you were trying to explain to me.

 

[Dr.Peterson]

Thank you for your help! Based upon what you said, this is what I came up with. Again, please let me know if I applied what you were trying to explain to me.

Your graph isn't quite right; they don't intersect at (-1,3) and (1,3). Also your inequality below that is wrong! Which graph is higher near 0?

The second version has the right inequality, though you need to state the condition under which it is true; but why do you show the left limit for one function and the right limit for the other?? The way you applied the squeeze theorem the first time was better.

 

[Yoyoma2732]

What do you mean by state the condition of the inequality. Would that be 1+x^4 <1 and (2x^2)+1 > 1?

 

[Dr.Peterson]

What do you mean by state the condition of the inequality. Would that be 1+x^4 <1 and (2x^2)+1 > 1?

The inequality you stated isn't always true! At least restate their hint, that this is true when x is between -1 and 1. (It's really true on a somewhat larger interval, and ideally you would prove that.)

What you say here is wrong. Why would you say that?

Apart from this, your new work is correct. Just one detail: what you show as the graph of f(x) is not; f(x) is actually the other two graphs combined, since each point on the graph of f(x) is on one or the other, not somewhere in between. But your graph represents the idea of the squeeze theorem.

 

[Yoyoma2732]

I wrote 1+x^4 <1 and (2x^2)+1 > 1 because I was confused about the interval. The attached photos show the changes that I made.

 

[Dr.Peterson]

I wrote 1+x^4 <1 and (2x^2)+1 > 1 because I was confused about the interval. The attached photos show the changes that I made.

No! How can you replace x with 1+x^4 and think it means the same thing? In fact, these two inequalities are entirely false. Did you look at the graph to check them?

As for what I did mean, look at your graph. Unfortunately, your graph is not accurate. Try solving to find when 1+x^4 = 1+2x^2, and you'll see where the two graphs really intersect.

Once you have it right, you'll see that for x further than 2 away from 0, the order of the two functions reverses; that's why you need to state the condition under which this one is less than that one.



This is nonsense! How can you redefine the function you were given, and do so with the same domain for both pieces? This is not where -1<x<1 belongs.



No, this change makes the inequality NEVER true! Don't you realize that f(x) is always equal to one polynomial or the other? It is never between them.

I think you've turned off your brain and tried to blindly follow what I said without trying to understand it.

 

[Yoyoma2732]

Dr. Peterson,

With this online learning, there is a disconnect between student and the teacher. I cannot ask questions in my class when I am unsure and my teacher is unavailable quite often. I have been researching and trying to teach myself these brand new concepts for countless hours. These topics are very challenging for me and I would hope you would have some patience with me while I try to figure out calculus. I am trying to follow your advise since you have years of experience tutoring students and it is frustrating that you think my brain is turned off as I am doing my best to try to follow your wisdom.

Thank you for your help.

 

[Dr.Peterson]

I know I was too harsh, but it's frustrating on my end, after you got all the main ideas right, to see you saying things that are clearly not true, and really do look like you aren't thinking about what you write. Checking is something that I always emphasize when I teach, because it's the only way to avoid mistakes. As I tell students, "Think. Write what you thought, in detail. Think about what you wrote. Then fix it."

I should confess that what I said about being further than 2 was wrong; that should have been [imath]\sqrt{2}[/imath]. I had a slightly wrong graph in my mind. Here is the right graph:



The important thing is that, taking their hint, when x is between -1 and 1 (we don't need to worry about the exact interval, only about some region near 0), the blue curve is above the red. And that's all I wanted you to add to your proof:

When [imath]-1<x<1[/imath], [imath]1+x^4\le f(x)\le 1+2x^2[/imath], ...​

Under that condition, the inequality is true. Without it, it isn't true.

 

[Yoyoma2732]

Dr. Peterson,

I want you to know that I would never purposefully try to waste your time or my time. I understand where you are coming from and I appreciate your help on this problem. Thank you again.

 

[lex]

Another couple of ideas for proofs:
[imath]\text{}\\ \hspace4ex 1≤f(x)≤ \text{Max\{}1+x^4,1+2x^2\text{\}} \hspace2ex[/imath]

[imath]\hspace4ex[/imath] Therefore by the 'Squeeze Theorem' [imath]\lim \limits_{x \to 0} f(x)=1 \hspace16ex[/imath](As [imath]x \to 0[/imath], both [imath]1+x^4 \to 1[/imath] and [imath]1+2x^2 \to 1[/imath],
[imath]\hspace65ex[/imath]therefore so does the greater of the two of them).








Or

When |x|<1,

[imath]\; x^4≤2x^2 \hspace2ex[/imath] (since [imath]x^2(x^2-2)≤0[/imath])
[imath]\therefore 1+x^4<1+2x^2[/imath]
[imath]\therefore 1≤f(x)≤1+2x^2 \hspace2ex[/imath] when |x|<1
Therefore by the 'Squeeze Theorem' [imath]\lim \limits_{x \to 0} f(x)=1[/imath]

 

[lex]

When |x|<1,

[imath]\; x^4≤2x^2 \hspace2ex[/imath]
[imath]\therefore 1+x^4[/imath] ≤ [imath] 1+2x^2[/imath]
...for the sake of correctness!

 

[Steven G]

x^4 >= 0, so how can 1+x^4 <1???

 

[Steven G]

If x is between -1 and 1, then we write -1<x<1 (Not that x is written in between -1 and 1)
If you write that -1 < 1+2x^2 < 1, then 1+2x^2 is between -1 and 1. Got it?