# help with squeeze theorem.

• Jan 7th 2008, 10:26 PM
rcmango
help with squeeze theorem.
problem by squeeze theorem: 0<=sin^2(x)/x^4 <=1

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

I believe the bottom denominator, x^4 continues on to infinity. it just keeps on going until infinity, so a 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.

is this correct?
• Jan 7th 2008, 10:33 PM
Jhevon
Quote:

Originally Posted by rcmango
problem by squeeze theorem: 0<=sin^2(x)/x^4 <=1

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

I believe the bottom denominator, x^4 continues on to infinity. it just keeps on going until infinity, so a 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.

is this correct?

this is not correct, what you're trying to prove is false. say x = 1/2, then we have $\frac {\sin^2 x}{x^4} > 1$. i am thinking of x in radians here. if you want to do it in degrees, a smaller number will suffice, say x = 0.00001
• Jan 8th 2008, 01:46 AM
Quote:

problem by squeeze theorem: 0<=sin^2(x)/x^4 <=1

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

I believe the bottom denominator, x^4 continues on to infinity. it just keeps on going until infinity, so a 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.

is this correct?
Looks good to me. You might want to set it out a bit more formally though. Start with something like:
$\frac {0}{x^4} \leq \frac {\sin ^2(x)}{x^4} \leq \frac {1}{x^4}$ for all x, then take the limit for all 3 parts and conclude that the limit you are looking at is between 0 and 0, and therefore is 0.
• Jan 8th 2008, 10:10 AM
Jhevon
Quote:

$\frac {0}{x^4} \leq \frac {\sin ^2(x)}{x^4} \leq \frac {1}{x^4}$ for all x

well, that makes sense. but before when we had 1 instead of 1/x^4, it didn't
• Jan 8th 2008, 04:38 PM
Quote:

well, that makes sense. but before when we had 1 instead of 1/x^4, it didn't
I know what you mean. I often don't understand people when they don't say exactly what they mean, but this time I had the advantage of just having found out that squeeze theorem is another name for the sandwich rule.
• Jan 8th 2008, 04:42 PM
Jhevon
Quote: