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Master_of_the_Faster
09-3-2008, 10:16 PM
I remembered one of the problems from my first unit test in calculus and would like to share it with you guys. It basically went like this: use the sandwich theorem to find the limit of
5-(x^2)(sin(1/x)). I know that there are two functions that surround this function when dealing with the sandwich theorem and that in an example, I saw that the function was basically sandwiched between -x^2 and x^2. Does anyone know which functions
5-(x^2)(sin(1/x)) is sandwiched between and why?

Edit: Sorry that I didn't mention that this is high school math (well both semesters of calculus in college, but taught in high school) in the title of the thread.

dooty_7
09-3-2008, 11:30 PM
I haven't seen the Sandwich theorem in a few years so I am sorry if I am rusty,

The following picture shows the function sandwiched between x^2 + 5 and -(x^2) + 5. Therefore since the limit as x->0 for the upper and lower functions is 5, then the limit of the original function is also 5.

PS if this is all wrong, then someone please correct me!
EDIT: Also if you need any clarification, just post and ask away

Icenri
09-4-2008, 12:36 AM
You're right dooty but I'll expand a little bit more the explanation because he might not get clear the "why":

When using sandwich theorem always remember that |Sin(whatever)| <= 1
so (f)(Sin(whatever)) will be sandwiched by f and -f. f is here x^2 and whatever is here 1/x. Add the 5 at the end in both sandwich functions. The limit is clear since for both sandwich functions, f(0) = 5

PS: You should have pointed out that the limit had x tending to 0.

Master_of_the_Faster
09-4-2008, 01:10 AM
Ah yes, I forgot to point out that x is approaching zero when dealing with this limit, but I'm glad you guys could help. I just read over the sandwich theorem stuff online and on a CD that I had for a while (since 6th grade). While I understand the algebra behind this problem, I'm not sure why I'm not able to graph 5-(x^2)(sin(1/x)) the way it was indicated by dooty.

Edit: Nevermind, I think I just needed to zoom in or something.

igotrhythm
09-4-2008, 07:46 PM
If I recall correctly, the Squeezing Theorem goes something like this:
If g(x) <= f(x) <= h(x) for all values of x,
and if for a specific value of x, called n here, g(n) = h(n),
then g(n) = f(n) = h(n).

Might have a book dealing with it around here somewhere, but meh, a picture really is worth a thousand words in this case.

You really need the "sandwich" going here, because if I recall correctly, lim x->0 (sin (1/x)) doesn't exist because the period gets shorter and shorter as you approach 0.

Nightfirecat
09-23-2008, 05:14 PM
Yeah, the limit shouldn't exist, since the sandwich theory tends to lead to oscillating fuctions which don't have any connections between points.