# Squeeze theorem.

1. Jan 6, 2008

### 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

2. Jan 6, 2008

### olgranpappy

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

3. Jan 6, 2008

### Dick

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

4. Jan 6, 2008

### olgranpappy

:rofl:

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

5. Jan 6, 2008

### olgranpappy

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

which would not have the same limit....

6. Jan 6, 2008

### Dick

Let's not forget sin(2^x).

7. Jan 6, 2008

### olgranpappy

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

8. Jan 6, 2008

9. Jan 6, 2008

### olgranpappy

yes, you are on the right track.

10. Jan 6, 2008

### Dick

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

11. Jan 6, 2008

### 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.

12. Jan 6, 2008

### olgranpappy

it is, indeed, never greater than 1.

no... it's always more than -1... it is always the square of a real number...

13. Jan 6, 2008

### rcmango

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

14. Jan 6, 2008

### Dick

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

15. Jan 6, 2008

### Gib Z

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?

16. Jan 7, 2008

### 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.

17. Jan 8, 2008

### olgranpappy

you must mean to say
$$lim_{x\to\infty}\sin^2(x)/x^4=0$$
, right?

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.

18. Jan 8, 2008

### Gib Z

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: Jan 8, 2008