Answer
$\displaystyle \lim_{x\rightarrow 3^{-}}\frac{1}{x-3}=-\infty,\quad \displaystyle \lim_{x\rightarrow 3^{+}}\frac{1}{x-3}=+\infty$.
Work Step by Step
Approaching 3 from the left, ($x\rightarrow 3^{-}$)
x is a positive number slightly less than 3, so the denominator becomes smaller and smaller in absolute value, nearing zero, and is negative.
The quotient represents the reciprocal of extremely small negative numbers,
which are negative numbers with extremely large absolute values.
Thus, $\displaystyle \lim_{x\rightarrow 3^{-}}\frac{1}{x-3}=-\infty$.
Approaching from the right, ($x\rightarrow 3^{+}$)
$x $ is slightly larger than 3, so this time, the denominator remains positive.
The quotient represents the reciprocal of extremely small positive numbers,
which are extremely large positive munbers
Thus, $\displaystyle \lim_{x\rightarrow 3^{+}}\frac{1}{x-3}=+\infty$.
