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.

• you can't say that a limit is oscillating, the limit doesn't exist – Nick Pavlov Oct 1 '17 at 15:52
• @NickPavlov I know but that is what Mathematica gave me, I am just trying to show work for how I can reach the conclusion of how the limit does not exist using algebraic techniques. – EnlightenedFunky Oct 1 '17 at 15:57

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 $\sin(\frac{1}{u_n - \pi}) \to_{n \to \infty} \sin(y)$

• Exactly! And if you haven't seen definition of limits using sequences, you can use this argument to say that there exists epsilon such that for every delta there exists x such that etc etc etc... – user334639 Oct 1 '17 at 21:15

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.

• You cannot split the limit to the product unless each limit alone exists and is finite. – user334639 Oct 1 '17 at 21:13
• @user334639 Your right, but why is that? Because how then would $\lim_{x\to0}(\sin(\frac{1}{x}))$ be computed? – EnlightenedFunky Oct 1 '17 at 21:45
• The limit can't be computed. It can be shown not to exist. The accepted answer shows that, by finding sequences that produce different limits as $x_n\to\pi$. – user334639 Oct 1 '17 at 23:46