# Math Help - Prove this limit

1. ## 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.

$
\lim_{x \to 0^{+}}$
$\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

2. Hello,
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.

$
\lim_{x \to 0^{+}}$
$\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
$\forall t \in \mathbb{R} ~,~ -1 \le \sin(t) \le 1$
What if you let $t=\frac \pi x$ ?

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

3. Originally Posted by Moo
Hello,

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

Then compose with the exponential function (which is increasing) and then multiply by $\sqrt{x}$ (which is positive).
Not gonna lie I have no idea what the first part of that math equation says

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

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

but that is as far as I can understand

4. The first part just means sin(t) has the domain of all real numbers

We also know

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

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

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

Can you do the next step?

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

We also know

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

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

$e^{-1} \le e^{sin(t)} \le e^{1}$
and then substitute "t" with $\pi/x$ ?

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

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

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

7. Because the sine function is bounded by -1 below and 1 above we have:
$e^{ - 1} \le e^{\sin \left( {\frac{\pi }{x}} \right)} \le e$.

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

But $\lim _{x \to 0^ + } \sqrt x = 0$ so the limit is 0.

8. 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

$

\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

9. Yep

10. so are the two squeezing functions always going to resemble the function they are squeezing?

in this case, the middle function had $\sqrt{x}$ and "e" in it, so we made the squeezing functions to look like this also

11. 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.

12. 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?

13. 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: $\lim _{x \to 0} \left| x \right|\left( {\arctan \left( {\frac{\pi }{x}} \right)} \right)$.
Because $- \frac{\pi }{2} \le \arctan (x) \le \frac{\pi }{2}$ we have $- \frac{\pi }{2}\left| x \right| \le \left| x \right|\arctan \left( {\frac{\pi }{x}} \right) \le \left| x \right|\frac{\pi }{2}$.

Now the limit is clear.

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

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 $-\pi/2$ and $\pi/2$

And that's pretty much as far as the resemblance between the original and squeezing functions go

15. Here is an example

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

You could use squeeze thoerem for this.

$f(x) = 0$

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

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

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

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