# Prove this limit

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• Sep 15th 2008, 01:45 PM
silencecloak
Prove this limit
We just learned this (well, a much more basic version) today in calculus, and I have no idea where to start on this problem.

$\displaystyle \lim_{x \to 0^{+}}$
$\displaystyle \sqrt{x}*e^{\sin(\pi/x)} = 0$

"
your task is to first find two outer functions that "squeeze" the given one to the right of zero; then compute the appropriate limits and make your conclusion"

EDIT:

http://www.mathhelpforum.com/math-he...e-theorem.html

i searched and found the above thread (which made me laugh because it's the same problem!), but im still very lost

thanks
• Sep 15th 2008, 01:55 PM
Moo
Hello,
Quote:

Originally Posted by silencecloak
We just learned this (well, a much more basic version) today in calculus, and I have no idea where to start on this problem.

$\displaystyle \lim_{x \to 0^{+}}$
$\displaystyle \sqrt{x}*e^{\sin(\pi/x)} = 0$

"
your task is to first find two outer functions that "squeeze" the given one to the right of zero; then compute the appropriate limits and make your conclusion"

EDIT:

http://www.mathhelpforum.com/math-he...e-theorem.html

i searched and found the above thread (which made me laugh because it's the same problem!), but im still very lost

thanks

$\displaystyle \forall t \in \mathbb{R} ~,~ -1 \le \sin(t) \le 1$
What if you let $\displaystyle t=\frac \pi x$ ?

Then compose with the exponential function (which is increasing) and then multiply by $\displaystyle \sqrt{x}$ (which is positive).
• Sep 15th 2008, 02:03 PM
silencecloak
Quote:

Originally Posted by Moo
Hello,

$\displaystyle \forall t \in \mathbb{R} ~,~ -1 \le \sin(t) \le 1$
What if you let $\displaystyle t=\frac \pi x$ ?

Then compose with the exponential function (which is increasing) and then multiply by $\displaystyle \sqrt{x}$ (which is positive).

Not gonna lie I have no idea what the first part of that math equation says

I do understand that $\displaystyle -1 \le \sin(t) \le 1$

and that this is true even while
$\displaystyle t=\frac \pi x$

but that is as far as I can understand
• Sep 15th 2008, 02:05 PM
11rdc11
The first part just means sin(t) has the domain of all real numbers

We also know

$\displaystyle -1 \le \sin(t) \le 1$

So make it look like your original function by putting the number e in

$\displaystyle e^{-1} \le e^{sin(t)} \le e^{1}$

Can you do the next step?
• Sep 15th 2008, 02:14 PM
silencecloak
Quote:

Originally Posted by 11rdc11
The first part just means sin(t) has the domain of all real numbers

We also know

$\displaystyle -1 \le \sin(t) \le 1$

So make it look like your original function by putting the number e in

$\displaystyle e^{-1} \le e^{sin(t)} \le e^{1}$

and then substitute "t" with $\displaystyle \pi/x$ ?

and that's it?

but then i only want the lim as x tends to 0 from the right, so i would leave off the whole -1 side

thats not to bad, just have to wrap my head around it
• Sep 15th 2008, 02:16 PM
Moo
Quote:

Originally Posted by silencecloak
and then substitute "t" with $\displaystyle \pi/x$ ?

and that's it?

thats not to bad, just have to wrap my head around it

Yes, since it is okay for any value of t :)
I used t so that you could see more easily, but I think it rather confused you (Worried)
• Sep 15th 2008, 02:17 PM
Plato
Because the sine function is bounded by -1 below and 1 above we have:
$\displaystyle e^{ - 1} \le e^{\sin \left( {\frac{\pi }{x}} \right)} \le e$.

So we have this $\displaystyle \sqrt x e^{ - 1} \le \sqrt x e^{\sin \left( {\frac{\pi }{x}} \right)} \le \sqrt x e$.

But $\displaystyle \lim _{x \to 0^ + } \sqrt x = 0$ so the limit is 0.
• Sep 15th 2008, 02:24 PM
silencecloak
ok so for the grand finally, my final answer is

"your task is to first find two outer functions that "squeeze" the given one to the right of zero; then compute the appropriate limits and make your conclusion"

the two outer functions are

$\displaystyle \sqrt x e^{ - 1} \le \sqrt x e^{\sin \left( {\frac{\pi }{x}} \right)} \le \sqrt x e^{1}$

and the limits of each function, lets call them h(x), f(x), g(x) all = 0
• Sep 15th 2008, 02:28 PM
11rdc11
Yep(Clapping)
• Sep 15th 2008, 02:32 PM
silencecloak
so are the two squeezing functions always going to resemble the function they are squeezing?

in this case, the middle function had $\displaystyle \sqrt{x}$ and "e" in it, so we made the squeezing functions to look like this also
• Sep 15th 2008, 02:40 PM
11rdc11
Not necessarily. I can't think of an example off the top of my head but I know when working with sequences and using the squeeze theorem it doesn't necessarily have to resemble the function. Let me see if I come up with an example.
• Sep 15th 2008, 02:57 PM
silencecloak
Quote:

Originally Posted by 11rdc11
Not necessarily. I can't think of an example off the top of my head but I know when working with sequences and using the squeeze theorem it doesn't necessarily have to resemble the function. Let me see if I come up with an example.

Might it be easier to just explain why in this case the two other functions did resemble the original in this way?
• Sep 15th 2008, 03:04 PM
Plato
Quote:

Originally Posted by silencecloak
so are the two squeezing functions always going to resemble the function they are squeezing?

That would be nice but not necessary.
Consider: $\displaystyle \lim _{x \to 0} \left| x \right|\left( {\arctan \left( {\frac{\pi }{x}} \right)} \right)$.
Because $\displaystyle - \frac{\pi }{2} \le \arctan (x) \le \frac{\pi }{2}$ we have $\displaystyle - \frac{\pi }{2}\left| x \right| \le \left| x \right|\arctan \left( {\frac{\pi }{x}} \right) \le \left| x \right|\frac{\pi }{2}$.

Now the limit is clear.
• Sep 15th 2008, 03:18 PM
silencecloak
Quote:

Originally Posted by Plato
That would be nice but not necessary.
Consider: $\displaystyle \lim _{x \to 0} \left| x \right|\left( {\arctan \left( {\frac{\pi }{x}} \right)} \right)$.
Because $\displaystyle - \frac{\pi }{2} \le \arctan (x) \le \frac{\pi }{2}$ we have $\displaystyle - \frac{\pi }{2}\left| x \right| \le \left| x \right|\arctan \left( {\frac{\pi }{x}} \right) \le \left| x \right|\frac{\pi }{2}$.

Now the limit is clear.

Ok, so at least the way I'm seeing it, the two squeezing functions still do in a way relate to the original. In the way that sin relates to -1 and 1 arctan relates to $\displaystyle -\pi/2$ and $\displaystyle \pi/2$

And that's pretty much as far as the resemblance between the original and squeezing functions go
• Sep 15th 2008, 03:29 PM
11rdc11
Here is an example

$\displaystyle \lim_{x \to \infty} \frac{1}{e^{x}}$

You could use squeeze thoerem for this.

$\displaystyle f(x) = 0$

$\displaystyle g(x) = \frac{1}{e^x}$

$\displaystyle h(x) = \frac{1}{x}$

Prob not the best example but you can see what I mean lol
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