# Squeeze theorem.

## Homework Statement

evaluate the limit.

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

squeeze theorem.

## 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

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

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

which would not have the same limit....

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

which would not have the same limit....

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

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

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 :)

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olgranpappy
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...

yes, you are on the right track.

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

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
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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.

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

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

Gib Z
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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?

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
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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
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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.

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