Another Squeeze Theorem Question

**joatmon** · January 22nd 2011, 11:19 AM

OK, I think that I have this Squeeze Theorem figured out. Can somebody please look at this to tell me if I do?

Proove: $\displaystyle \lim_{x\to 0 } \sqrt{x^3+x^2}sin \frac{\pi}{x}=0}$

Let:
$f(x)=- \sqrt{x^3+x^2}$
$g(x)= \sqrt{x^3+x^2}sin \frac{\pi}{2}$
$h(x)= \sqrt{x^3+x^2}$

I think that I can state: $-1 \leq sin \frac{\pi}{x} \leq 1$

Then, multiplying each term by $\sqrt{x^3+x^2}$ :

$- \sqrt{x^3+x^2} \leq \sqrt{x^3+x^2}sin \frac{\pi}{2} \leq \sqrt{x^3+x^2}$

Since $\displaystyle \lim_{x\to 0 } f(x) = \displaystyle \lim_{x\to 0 } h(x) =0}$ , using Squeeze Theorem, I have proved that:

$\displaystyle \lim_{x\to 0 } \sqrt{x^3+x^2}sin \frac{\pi}{x}=0}$

Did I do this correctly? I appreciate anybody who takes a look at this for me.

**dwsmith** · January 22nd 2011, 11:49 AM

Originally Posted by joatmon

OK, I think that I have this Squeeze Theorem figured out. Can somebody please look at this to tell me if I do?

Proove: $\displaystyle \lim_{x\to 0 } \sqrt{x^3+x^2}sin \frac{\pi}{x}=0}$

Let:
$f(x)=- \sqrt{x^3+x^2}$
$g(x)= \sqrt{x^3+x^2}sin \frac{\pi}{2}$
$h(x)= \sqrt{x^3+x^2}$

I think that I can state: $-1 \leq sin \frac{\pi}{x} \leq 1$

Then, multiplying each term by $\sqrt{x^3+x^2}$ :

$- \sqrt{x^3+x^2} \leq \sqrt{x^3+x^2}sin \frac{\pi}{2} \leq \sqrt{x^3+x^2}$

Since $\displaystyle \lim_{x\to 0 } f(x) = \displaystyle \lim_{x\to 0 } h(x) =0}$ , using Squeeze Theorem, I have proved that:

$\displaystyle \lim_{x\to 0 } \sqrt{x^3+x^2}sin \frac{\pi}{x}=0}$

Did I do this correctly? I appreciate anybody who takes a look at this for me.

Looks good.

**dwsmith** · January 22nd 2011, 11:55 AM

This had an unique graph.

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