# Showing a limit does not exist?

• Jul 20th 2012, 07:20 AM
LiberalArtMajorTakingCalc
Showing a limit does not exist?
$\lim_{x \to 2 } \frac{3}{(x - 2)}$

I cheated by looking at a graph and it goes to $\pm \infty$, but when I try to solve it algbraically by substitution, I plug $2$ in and get undefined in the denominator, as I'd expect with an infinite limit. But how do I show the algebraic work that the behavior of the function as the $\lim_{x \to 2 }$ goes to both $\pm \infty$?
• Jul 20th 2012, 07:41 AM
Plato
Re: Showing a limit does not exist?
Quote:

Originally Posted by LiberalArtMajorTakingCalc
$\lim_{x \to 2 } \frac{3}{(x - 2)}$

I cheated by looking at a graph and it goes to $\pm \infty$, but when I try to solve it algbraically by substitution, I plug $2$ in and get undefined in the denominator, as I'd expect with an infinite limit. But how do I show the algebraic work that the behavior of the function as the $\lim_{x \to 2 }$ goes to both $\pm \infty$?

Let $f(x)=\frac{3}{(x - 2)}$ define $x_n=2+\frac{1}{n}~~y_n=2-\frac{1}{n}$

It is clear that $x_n\to 2^+~\&~y_n\to 2^-$, but $f(x_n)=3n\to\infty~\&~f(y_n)=-3n\to-\infty$.
• Jul 20th 2012, 08:07 AM
LiberalArtMajorTakingCalc
Re: Showing a limit does not exist?
Where did $2+\frac{1}{n}$ and $2-\frac{1}{n}$ come from?