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

Apr 11, 2016

As $x$ approaches $3$ from the right, the numerator is a positive number and the denominator is also positive.

Numerator $\rightarrow$ a positive
Denominator $\rightarrow$ $0$ through positive values,

So the ratio increases without bound.

We write ${\lim}_{x \rightarrow {3}^{+}} \frac{2}{x - 3} = \infty$.

Although not in widespread use (I think), it may be helpful to write:
${\lim}_{x \rightarrow {3}^{+}} \frac{2}{x - 3}$ has form $\frac{+}{0} ^ +$, which leads to ${\lim}_{x \rightarrow {3}^{+}} \frac{2}{x - 3} = \infty$.

Bonus

${\lim}_{x \rightarrow {3}^{+}} \frac{- 5}{x - 3}$ has form $\frac{-}{0} ^ +$, which leads to ${\lim}_{x \rightarrow {3}^{+}} \frac{- 5}{x - 3} = - \infty$
${\lim}_{x \rightarrow {3}^{-}} \frac{2}{x - 3}$ has form $\frac{+}{0} ^ -$, which leads to ${\lim}_{x \rightarrow {3}^{-}} \frac{2}{x - 3} = - \infty$
${\lim}_{x \rightarrow {3}^{+}} \frac{x - 1}{x - 3}$ has form $\frac{+}{0} ^ +$, which leads to ${\lim}_{x \rightarrow {3}^{+}} \frac{x - 1}{x - 3} = \infty$
${\lim}_{x \rightarrow {3}^{+}} \frac{x - 7}{x - 3}$ has form $\frac{-}{0} ^ +$, which leads to ${\lim}_{x \rightarrow {3}^{+}} \frac{x - 7}{x - 3} = - \infty$