Math Option - Calculus Squeeze Theorem

[Dax]

Hi Guys,

I have another doubt, this time in Paper 3, Squeeze Theorem.

I have a question

f(x) = x^3 sin(pi/x)cos(pi/x)

lim f(x)

x -> 0

How do you solve this question, and in general more questions like this through Squeeze Theorem?

Thanks!

[flinquinnster]

Hey! I myself never really liked using squeeze theorem and just avoided using it, but I might be able to provide some help. Generally, to use squeeze theorem to find the limit of a function g(x) as x -> a, then you attempt to find functions f(x) and h(x) where f(x) < g(x) < h(x) (technically lesser than or equal to signs) and the Lim[x->a]f(x) = Lim[x->a]h(x). Whilst that probably doesn't make much more sense than the textbook, I can't really think of a better way to explain it - you just need to find 2 other functions that sandwich your given function, and then show how these 2 functions have the same limit and thus your given function has the same limit as it must be 'squeezed' in between the limits of the 2 other functions.

So here, as sin(pi/x) and cos(pi/x) have an absolute value lesser than or equal to 1, 0 < x^3sin(pi/x)cos(pi/x) < x^3 (again, where those are lesser than or equal to signs). Now, the limit of 0 as x-> 0 is 0. The limit of x^3 as x->0 is again 0. Thus, using squeeze theorem, the limit of x^3sin(pi/x)cos(pi/x) is 0. If you're still unsure, if you input the function and find the y-value for a small x-value like 0.001, then you can approximately see that the function does indeed approach 0.

Apologies if the above isn't very clear - I can't figure out how to do all the nice mathematical symbol formatting and have completely forgotten how to upload screenshots.

EDIT: I only just realised that the inequality I created works under the assumption that x > 0 and that each function has absolute value signs around it, which is definitely not necessarily the case. As I mentioned I actually disliked using squeeze theorem so I'm not entirely sure how else to approach the question using squeeze theorem. Sorry for not being that much help, but maybe if I leave my own attempt up here someone might be able to contribute a better solution!

Edited March 23, 2014 by flinquinnster

[Dax]

The answer on the back of the book says -1. I multiplied and divided the function by 2 so the function became a variant sin2x the used the same. I too get 0 as the limit but it seems to be wrong.

[maroctam]

Wolframalpha also says that the limit is 0, so I'm guessing that the -1 in your book is a mistake.

Does the question explicitly say "Use the squeeze theorem to evaluate this limit"? Because, like you guys, I don't like the squeeze theorem either and I think the easier way to approach this question is by writing u=pi/x and then replacing all the xs by us giving

lim (u->∞) (pi^3 sin(u)cos(u)) / u^3

and since pi^3 sin(u)cos(u) is bounded (by (-pi^3)/2 and (pi^3)/2, I think) and u^3 approaches infinity, the limit is 0.

I seriously don't know how you would do this using the squeeze theorem, so I'm sorry for not being helpful in that aspect!

[Dax]

I got how to do it using Squeeze Theorem,

You can Multiply and Divide it by 2, to get the expansion of Sin 2(pi/x).

Then you know Sin 2(pi/x) is bound between -1 and 1

multiply all three sides of inequality by (x^3)/2.
That should given you a bounded region. I got 0 as well. So the book is wrong then eh?