# Squeeze theorem.

#### rcmango

1. The problem statement, all variables and given/known data

evaluate the limit.

lim x -> infinity sin2^x / x^4

2. Relevant equations

squeeze theorem.

3. The attempt at a solution

I believe the squeeze theorem could be relevant here.
also, i know that sin 1 is 0. so couldn't this problem be something like 0 < sinx/ x < 0

thus also making the answer 0? am i on the right track, please help, thanks.

Related Calculus and Beyond Homework News on Phys.org

#### olgranpappy

Homework Helper
it's not clear what you mean by sin2^x

#### Dick

Homework Helper
it's not clear what you mean by sin2^x
Not this again! This time I'll guess (sin(x))^2. Am I right?

#### olgranpappy

Homework Helper
Not this again! This time I'll guess (sin(x))^2. Am I right?

:rofl:

I'm going to guess sin(x^2)... let's see who's right... Of course, the answer is the same either way.

#### olgranpappy

Homework Helper
... unfortunately he could also mean ( sin(2) )^x

which would not have the same limit....

#### Dick

Homework Helper
... unfortunately he could also mean ( sin(2) )^x

which would not have the same limit....
Let's not forget sin(2^x).

#### olgranpappy

Homework Helper
... so we can say that there is a 75 percent chance that the answer is zero, and a 25 percent chance it's infinity.

#### rcmango

okay sorry for the ambiguity. I'll try harder to make this more clear next time, i truly did not know the correct form of parenthesis for this one: here it is, in image form:
http://img244.imageshack.us/img244/4076/72107252du9.jpg [Broken]

there ya go :)

Last edited by a moderator:

#### olgranpappy

Homework Helper

...
thus also making the answer 0? am i on the right track, please help, thanks.

yes, you are on the right track.

#### Dick

Homework Helper
I win! That's sin(x)^2. So squeeze it. -1<=sin(x)<=1. What about sin(x)^2?

#### rcmango

okay, so since i know that sin is limited to 1 and -1, then also sin(x)^2 is also limited to 1?

so which is it then, is sin(x)^2 1 or -1? i know its no more or no less than either of the two.

thanks so far.

#### olgranpappy

Homework Helper
okay, so since i know that sin is limited to 1 and -1, then also sin(x)^2 is also limited to 1?
it is, indeed, never greater than 1.

so which is it then, is sin(x)^2 1 or -1? i know its no more or no less than either of the two.
no... it's always more than -1... it is always the square of a real number...
thanks so far.

#### rcmango

so then it is fair to say, it is 1.

#### Dick

Homework Helper
It's not 1. 0<=sin^2(x)<=1. Now finish the problem.

#### Gib Z

Homework Helper
Come on rcmango, it doesn't even matter really about the inequality, you could have -243425254 <= sin^2 (x) <= 429837928562983462394 for all we care. The point is the numerator is bounded. Is the denominator? What does this mean?

#### rcmango

Okay, so we know that sin^2 (x) is bounded between 0 and 1,

the denominator, x^4 is not bounded. it just keeps getting bigger, so number could be like 1, 1/16, 1/81 and so on ... right? thus getting closer to 0. So sin^2 (x)/x^4 = 0.

I've put alot of effort thinking about this, i'm not very good at it, thanks for the help.

#### olgranpappy

Homework Helper
Okay, so we know that sin^2 (x) is bounded between 0 and 1,

the denominator, x^4 is not bounded. it just keeps getting bigger, so number could be like 1, 1/16, 1/81 and so on ... right? thus getting closer to 0. So sin^2 (x)/x^4 = 0.
you must mean to say
$$lim_{x\to\infty}\sin^2(x)/x^4=0$$
, right?

I've put alot of effort thinking about this, i'm not very good at it, thanks for the help.
It would also probably be useful to graph the function and you will see right away that as 'x' gets large
sin^2(x)/x^4 get very close to zero quite quickly.

#### Gib Z

Homework Helper
rcmango: Yes that is a reasonable manner to think in, but it would be better to have done it more like this;

Since we have the strict inequality $0 \leq \sin^2 x \leq 1$, we also have the inequality (provided x is not equal to zero) - $$0 \leq \frac{\sin^2 x}{x^4} \leq \frac{1}{x^4}$$. Now taking limits as we let the value of x be arbitrarily large, we arrive at:

$$0 \leq \lim_{x\to \infty} \frac{\sin^2 x}{x^4} \leq \lim_{x\to \infty} \frac{1}{x^4}$$. Since the limit on the right is also equal to 0, by the squeeze theorem (which your thread name implies was originally to be used), the limit $$\lim_{x\to \infty} \frac{\sin^2 x}{x^4}$$ is equal to 0.

Last edited:

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving