Is there an algebraic method to get at this limit? I know the limit of this function as it approaches $\pi$ is oscillating between $-1$, and $1$ meaning it does not exist.


  $$\lim_{x\to\pi}(\sin(\frac{1}{x-\pi}))$$


I was just wondering if graphing is considered an algebraic technique?  Because I don't know how to approach this algebraically.
I do not really get what you call algebraic methods (and non-algebraic methods), but to show that the function is oscillating between $-1$ and $1$, consider the following sequence (with $y$ being any real number)  :

$u_n = \frac{1}{y+2n\pi} + \pi$

You can easily check that $u_n \to_{n \to \infty} \pi$ and that $\sin(\frac{1}{u_n - \pi}) \to_{n \to \infty} \sin(y)$
A way to approach the limit is by using limit properties, squeeze theorem, and algebra techniques.


  $$\lim_{x\to\pi}(\sin(\frac{1}{x-\pi}))$$
  $$\lim_{x\to\pi}(\sin(\frac{1}{x-\pi})\times\frac{1}{\frac{x-\pi}{x-\pi}})$$
  $$\lim_{x\to\pi}(\frac{\sin(\frac{1}{x-\pi})}{\frac{1}{x-\pi}}\times\frac{1}{x-\pi})$$
  $$\lim_{x\to\pi}((x-\pi)(\sin{\frac{1}{x-\pi}})(\frac{1}{x-\pi})$$


Now applying limit properties:


  $$\lim_{x\to \pi}((x-\pi)(\sin(\frac{1}{x-\pi}))\times\lim_{x\to\pi}(\frac{1}{x-\pi})$$


Simplifying the limit on the left it can be done using squeeze theorem:


  $$f(x)\le g(x)\le h(x)$$
  $$-1\le \sin(\frac{1}{x-\pi})\le 1$$


Then,


  $$-(x-\pi)\le (x-\pi)\sin(\frac{1}{x-\pi})\le (x-\pi)$$


Afterwards a limit can be taken as x approaches $\pi$


  $$\lim_{x\to\pi}(-(x-\pi))\le \lim_{x\to\pi}((x-\pi)\sin(\frac{1}{x-\pi}))\le \lim_{x\to\pi}(x+\pi)$$


Which is equal to:


  $$0\le\lim_{x\to\pi}((x-\pi)\sin(\frac{1}{x-\pi}))\le 0$$


From this we can conclude that:


  $$(0)\times\lim_{x\to\pi}(\frac{1}{x-\pi})$$


Because the other limit has a similar form as:


  $$\lim_{x\to\pi}(\frac{1}{x-\pi})=\lim_{u\to0}(\frac{1}{u})$$


Therefore the limit is known to not exist therefore:


  $\lim_{x\to\pi}(\sin(\frac{1}{x-\pi}))$ does not exist.