squeeze theorem limit

**tedsauc** · Apr 1st 2012, 04:54 PM

Hello, I am trying to apply the squeeze theorem to the following:
e^sin(pi/x)
limit as x tends towards zero.

-1 < e^sin(pi/x) > 1

-(pi/x) < e^sin(pi/x) < (pi/x)

plugging in zero to the x variables gives me a limit of zero.

My question is, am I correct in how I have factored the (pi/x) components to the outer edges ?
Thank you.

**Prove It** · Apr 1st 2012, 05:01 PM

Originally Posted by tedsauc

Hello, I am trying to apply the squeeze theorem to the following:
e^sin(pi/x)
limit as x tends towards zero.

-1 < e^sin(pi/x) > 1

-(pi/x) < e^sin(pi/x) < (pi/x)

plugging in zero to the x variables gives me a limit of zero.

My question is, am I correct in how I have factored the (pi/x) components to the outer edges ?
Thank you.

While I agree that you could bound $\displaystyle \begin{align*} \sin{\left(\frac{\pi}{x}\right)} \end{align*}$ between $\displaystyle \begin{align*} -\frac{\pi}{x} \end{align*}$ and $\displaystyle \begin{align*} \frac{\pi}{x} \end{align*}$ , how could you possibly let $\displaystyle \begin{align*} x = 0 \end{align*}$ in the edges? You can't divide by 0... Exponentiating both sides won't help you either.

**tedsauc** · Apr 1st 2012, 05:04 PM

thanks. That is where I am stuck. I don't know what to put into the edges. If anyone can offer suggestions that would be most appreciated.

**Prove It** · Apr 1st 2012, 05:04 PM

Originally Posted by tedsauc

thanks. That is where I am stuck. I don't know what to put into the edges. If anyone can offer suggestions that would be most appreciated.

The fact is that the limit doesn't exist. limit Exp[Sin[Pi/x]] as x to 0 - Wolfram|Alpha

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