# Showing a limit does not exist?

• Jul 20th 2012, 06:20 AM
LiberalArtMajorTakingCalc
Showing a limit does not exist?
$\displaystyle \lim_{x \to 2 } \frac{3}{(x - 2)}$

I cheated by looking at a graph and it goes to $\displaystyle \pm \infty$, but when I try to solve it algbraically by substitution, I plug $\displaystyle 2$ 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 $\displaystyle \lim_{x \to 2 }$ goes to both $\displaystyle \pm \infty$?
• Jul 20th 2012, 06:41 AM
Plato
Re: Showing a limit does not exist?
Quote:

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

I cheated by looking at a graph and it goes to $\displaystyle \pm \infty$, but when I try to solve it algbraically by substitution, I plug $\displaystyle 2$ 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 $\displaystyle \lim_{x \to 2 }$ goes to both $\displaystyle \pm \infty$?

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

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