# How do you determine the limit of (2)/(x-3) as x approaches 3^-??

May 25, 2016

The expression $\frac{2}{x - 3}$ tends to negative infinity as $x$ approaches ${3}^{-}$ (that is, as $x \to 3$ while being always less than $3$)

#### Explanation:

Recall the definition of a limit:

We say that the limit of function $f \left(x\right)$ (defined for all $x$ in some neighborhood, however small, of value $x = a$, but not necessarily at point $x = a$ itself) is $A$ as $x \to a$
(written as ${\lim}_{x \to a} f \left(x\right) = A$ or $f {\left(x\right)}_{x \to a} \to A$)
if and only if for any $\epsilon > 0$ (measuring the closeness of $f \left(x\right)$ to value $A$)
there is such $\delta > 0$ (measure of the closeness of $x$ to value $a$) that
for any $x$ satisfying the inequality $| x - a | < \delta$
the following is true: $| f \left(x\right) - A | < \epsilon$

In other words, for any degree of closeness $\epsilon$ of $f \left(x\right)$ to $A$ we can find a distance $\delta$ of $x$ to $a$ such as the function $f \left(x\right)$ will be closer to $A$ than $\epsilon$ if $x$ is closer to $a$ than $\delta$.

With infinity the situation needs another definition, because infinity is not a number, like $A$ in the above definition.
We should supplement our definition for two cases of infinity - positive and negative.

We say that the limit of function $f \left(x\right)$ (defined for all $x$ in some neighborhood, however small, of value $x = a$, but not necessarily at point $x = a$ itself) is negative infinity
(written as ${\lim}_{x \to a} f \left(x\right) = - \infty$ or $f {\left(x\right)}_{x \to a} \to - \infty$)
if and only if for any real $A < 0$ (however large by absolute value)
there is such $\delta > 0$ (measure of the closeness of $x$ to value $a$) that
for any $x$ such as $| x - a | < \delta$
the following is true: $f \left(x\right) < A$

Similarly is defined the limit of positive infinity.

In our case, if $x \to {3}^{-}$ (that is, $x \to 3$ while $x < 3$), the fraction $\frac{2}{x - 3}$ is always negative.

Choose any $A < 0$ and resolve the inequality
$\frac{2}{x - 3} < A$
the solution is:
$x > 3 + \frac{2}{A}$
The larger by absolute value $A$ is (while being negative) - the closer $x$ should be to $3$ for our fraction to be less than $A$.
So, we always find that measure of closeness of $x$ to $3$ (while being smaller than $3$):
$\delta = \frac{2}{|} A |$
When $x$ is closer than this $\delta$ to $3$, the fraction will fall lower than $A$, no matter what negative $A$ we have chosen.