Evaluating Limit w/ Squeeze Theorem

  • Thread starter rcmango
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In summary, we can use the squeeze theorem to show that the limit of sin^2(x)/x^4 as x approaches infinity is equal to 0. This is because we can bound the function between 0 and 1, and as x gets larger, the denominator x^4 becomes unbounded, resulting in the limit being 0.
  • #1
rcmango
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Homework Statement



evaluate the limit.

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

Homework Equations



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

thus also making the answer 0? am i on the right track, please help, thanks.
 
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  • #2
it's not clear what you mean by sin2^x
 
  • #3
olgranpappy said:
it's not clear what you mean by sin2^x

Not this again! This time I'll guess (sin(x))^2. Am I right?
 
  • #4
Dick said:
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.
 
  • #5
... unfortunately he could also mean ( sin(2) )^x

which would not have the same limit...
 
  • #6
olgranpappy said:
... unfortunately he could also mean ( sin(2) )^x

which would not have the same limit...

Let's not forget sin(2^x).
 
  • #7
... 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
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 you go :)
 
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  • #9
rcmango said:

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


yes, you are on the right track.
 
  • #10
I win! That's sin(x)^2. So squeeze it. -1<=sin(x)<=1. What about sin(x)^2?
 
  • #11
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
rcmango said:
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.
 
  • #13
so then it is fair to say, it is 1.
 
  • #14
It's not 1. 0<=sin^2(x)<=1. Now finish the problem.
 
  • #15
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
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 a lot of effort thinking about this, I'm not very good at it, thanks for the help.
 
  • #17
rcmango said:
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
[tex]
lim_{x\to\infty}\sin^2(x)/x^4=0
[/tex]
, right?

I've put a lot 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.
 
  • #18
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 [itex] 0 \leq \sin^2 x \leq 1[/itex], we also have the inequality (provided x is not equal to zero) - [tex]0 \leq \frac{\sin^2 x}{x^4} \leq \frac{1}{x^4}[/tex]. Now taking limits as we let the value of x be arbitrarily large, we arrive at:

[tex]0 \leq \lim_{x\to \infty} \frac{\sin^2 x}{x^4} \leq \lim_{x\to \infty} \frac{1}{x^4}[/tex]. 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 [tex]\lim_{x\to \infty} \frac{\sin^2 x}{x^4}[/tex] is equal to 0.
 
Last edited:

What is the Squeeze Theorem?

The Squeeze Theorem, also known as the Sandwich Theorem, is a mathematical theorem that is used to evaluate limits of functions. It states that if two functions, f(x) and g(x), both have the same limit as x approaches a certain value c, and if there is a third function h(x) that is always between f(x) and g(x) near c, then h(x) also has the same limit as x approaches c.

How is the Squeeze Theorem used to evaluate limits?

The Squeeze Theorem is used to evaluate limits by finding two functions that have the same limit as the function being evaluated, and then finding another function that is always between those two functions. This third function will also have the same limit as the other two functions, and therefore, the limit of the original function can be determined using the limit of the third function.

What are the conditions for using the Squeeze Theorem?

The Squeeze Theorem can be used to evaluate limits only if the following conditions are met:

  • The two functions f(x) and g(x) have the same limit as x approaches c.
  • The third function h(x) is always between f(x) and g(x) near c.
  • The functions f(x) and g(x) are defined on an open interval containing c, except possibly at c itself.

Can the Squeeze Theorem be used to evaluate limits at infinity?

Yes, the Squeeze Theorem can also be used to evaluate limits at infinity. In this case, the third function h(x) should be bounded by two functions that approach the same value as x approaches infinity.

What are the benefits of using the Squeeze Theorem to evaluate limits?

The Squeeze Theorem is a powerful tool for evaluating limits because it allows us to determine the limit of a function without having to know the exact value of the function at the point being evaluated. It also provides a systematic approach for evaluating limits and can be used for both finite and infinite limits.

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