Thread: squeeze theorem limit

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.

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

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.

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&#91;Sin&#91;Pi&#47;x&#93;&#93; as x to 0 - Wolfram|Alpha