If the limit of a rational function produces a $$\frac{0}{0}$$ form...
- factor the numerator and denominator,
- divide out the common factor(s),
- then re-evaluate the limit .
Examples
The Limit Exists
Example 1
Evaluate $$\displaystyle \lim_{x\to-3}\frac{x^2+x-6}{x^2+8x+15}$$
Since the function is rational, we can try factoring both the numerator and denominator to identify common factors.
$$\begin{align*}% \lim_{x\to-3}\frac{x^2+x-6}{x^2+8x+15} % & = \lim_{x\to-3}\frac{{\color{blue}(x+3)}(x-2)}{{\color{blue}(x+3)}(x+5)} \\ % & = \lim_{x\to-3}\frac{x-2}{x+5} \end{align*} $$
Evaluate the simpler limit .
$$\displaystyle \lim_{x\to-3}\frac{x-2}{x+5} = \frac{-3-2}{-3+5} = -\frac 5 2$$
$$\displaystyle \lim_{x\to-3}\frac{x^2+x-6}{x^2+8x+15} = -\frac 5 2$$
The Limit Doesn’t Exist
Example 2
Evaluate: $$\displaystyle \lim_{x\to4}\frac{2x^2-7x-4}{x^3-8x^2+16x}$$
Since the function is rational, try factoring to find any common factors.
$$ \begin{align*} \lim_{x\to4}\frac{2x^2-7x-4}{x^3-8x^2+16x} & = \lim_{x\to4}\frac{{\color{blue}(x-4)}(2x+1)}{x{\color{blue}(x-4)}(x-4)} \\ % & = \lim_{x\to4}\frac{2x+1}{x(x-4)} \end{align*} $$
Evaluate the simpler limit .
$$\lim_{x\to4}\frac{2x+1}{x(x-4)} = \frac{2(4) + 1}{4(4-4)} = \frac 9 0$$
$$\displaystyle \lim_{x\to4}\,\frac{2x^2-7x-4}{x^3-8x^2+16x}$$ does not exist.
Practice Problems
Confirm the limit has an indeterminate form.
$$ \displaystyle \lim_{x\to2}\frac{x^2 + 5x -14}{x^2-4} = \frac{(2)^2+5(2)-14}{(2)^2-4}=\color{red}{ \frac 0 0} $$
Find and divide out any common factors.
$$ \begin{align*} \lim_{x\to2}\frac{x^2 + 5x -14}{x^2-4} & = \lim_{x\to2} \frac{(x+7){\color{blue}(x-2)}}{(x+2){\color{blue}(x-2)}} \\ % & = \lim_{x\to2} \frac{x+7}{x+2} \end{align*} $$
Evaluate the simpler limit .
$$\displaystyle \lim_{x\to2} \frac{x+7}{x+2} = \frac{2+7}{2+2} = \frac 9 4$$.
$$\displaystyle \lim_{x\to2}\frac{x^2 + 5x -14}{x^2-4} = \frac 9 4$$.
Confirm the limit has an indeterminate form.
$$ \displaystyle \lim_{x\to -1} \frac{x^2-4x-5}{x^2+10x+9} = \frac{(-1)^2-4(-1)-5}{(-1)^2+10(-1)+9} = \frac 0 0 $$
Find and divide out any common factors.
$$ \begin{align*} \lim_{x\to -1} \frac{x^2-4x-5}{x^2+10x+9} & = \lim_{x\to-1} \frac{(x-5){\color{blue}(x+1)}}{(x+9){\color{blue}(x+1)}} \\ % & = \lim_{x\to-1} \frac{x-5}{x+9} \\ \end{align*} $$
Evaluate the simpler limit .
$$\displaystyle \lim_{x\to-1} \frac{x-5}{x+9} = \frac{-1-5}{-1+9}=\frac{-6}{8} = -\frac 3 4$$.
$$\displaystyle \lim_{x\to -1} \frac{x^2-4x-5}{x^2+10x+9} = -\frac 3 4$$.
Confirm the limit has an indeterminate form.
$$ \displaystyle \lim_{x\to\frac 1 3} \frac{3x^2-7x+2}{3x^2+5x-2}% = \frac{3(\frac 1 3)^2-7(\frac 1 3) +2}{3(\frac 1 3) +5(\frac 1 3)-2}% = \color{red}{ \frac 0 0}% $$
Find and divide out any common factors.
$$ \begin{align*} \lim_{x\to\frac 1 3} \frac{3x^2-7x+2}{3x^2+5x-2} & = \lim_{x\to\frac 1 3} \frac{{\color{blue}(3x-1)}(x-2)}{{\color{blue}(3x-1)}(x+2)} \\ % & = \lim_{x\to\frac 1 3} \frac{x-2}{x+2} \end{align*} $$
Evaluate the simpler limit .
$$\displaystyle \lim_{x\to\frac 1 3} \frac{x-2}{x+2} = \frac{\frac 1 3 - 2}{\frac 1 3 + 2} = \frac{-\frac 5 3}{\frac 7 3} = -\frac 5 7$$.
$$\displaystyle \lim_{x\to\frac 1 3} \frac{3x^2-7x+2}{3x^2+5x-2} = - \frac 5 7$$.
Confirm the limit has an indeterminate form.
$$ \displaystyle \lim_{x\to-8} \frac{2x^2+19x+24}{2x^2+15x-8}% = \frac{2(-8)^2+19(-8)+24}{2(-8)^2+15(-8)-8} = \frac 0 0 $$
Find and divide out any common factors.
$$ \begin{align*} \lim_{x\to-8} \frac{2x^2+19x+24}{2x^2+15x-8} & = \lim_{x\to-8} \frac{{\color{blue}(x+8)}(2x+3)}{{\color{blue}(x+8)}(2x-1)} \\ % & = \lim_{x\to-8} \frac{2x+3}{2x-1} \end{align*} $$
Evaluate the simpler limit .
$$\displaystyle \lim_{x\to-8} \frac{2x+3}{2x-1} = \frac{2(-8)+3}{2(-8)-1} = \frac{-13}{-17} = \frac{13}{17}$$.
$$\displaystyle \lim_{x\to-8} \frac{2x^2+19x+24}{2x^2+15x-8} = \frac{13}{17}$$.
Confirm the limit has an indeterminate form.
$$ \displaystyle \lim_{x\to6} \frac{x^2-3x-18}{x^2-12x+36} = \frac{(6)^2-3(6)-18}{(6)^2-12(6)+36} = \color{red}{ \frac 0 0} $$
Find and divide out any common factors.
$$ \begin{align*} \lim_{x\to6} \frac{x^2-3x-18}{x^2-12x+36} & = \lim_{x\to6} \frac{(x+3){\color{blue}(x-6)}}{{\color{blue}(x-6)}(x-6)}\\ % & = \lim_{x\to6} \frac{x+3}{x-6} \end{align*} $$
Evaluate the simpler limit .
$$\displaystyle \lim_{x\to6} \frac{x+3}{x-6} = \frac{6 + 3}{6 -6} = \frac 9 0$$.
$$\displaystyle \displaystyle \lim_{x\to6} \frac{x^2-3x-18}{x^2-12x-36}$$ does not exist.
Confirm the limit has an indeterminate form.
$$ \begin{align*}% \displaystyle \lim_{x\to-5}\,% \frac{2x^2+13x+15}{x^3+10x^2+25x}% % & = \frac{2(-5)^2+13(-5)+15}{(-5)^3+10(-5)^2+25(-5)}% \\[6pt] % & = \frac{50 - 65 + 15}{-125 + 250 - 125}% \\[6pt] % & = \frac 0 0 \end{align*} $$
Find and divide out any common factors.
$$ \begin{align*} \lim_{x\to-5}\,% \frac{2x^2+13x+15} {x^3+10x^2+25x}% % & = \lim_{x\to-5}\,% \frac{% \blue{(x+5)}(2x+3)% } {% x\blue{(x+5)}(x+5)% } \\[6pt] % & = \lim_{x\to-5}\,% \frac{2x+3} {x(x+5)}% \end{align*} $$
Evaluate the simpler limit .
$$ \displaystyle\lim_{x\to-5}\,% \frac{2x+3}{x(x+5)}% % = \frac{2(-5) + 3}{-5(-5+5)}% % = \frac{-7} 0 $$.
$$ \displaystyle \lim_{x\to-5}\,% \frac{2x^2+13x+15}{x^3+10x^2+25x}% $$ does not exist.
Confirm the limit has an indeterminate form.
$$ \begin{align*}% \lim_{x\to -\frac 2 5}\,% \frac{15x^3+x^2-2x} {25x^2 + 20x + 4} % & = \frac{% 15\left(-\frac 2 5\right)^3+\left(-\frac 2 5\right)^2-2\left(-\frac 2 5\right)% } {% 25\left(-\frac 2 5\right)^2 + 20\left(-\frac 2 5\right) + 4 } \\[6pt] % & = \frac{% 15\left(-\frac 8 {125}\right)+\frac 4 {25}+\frac 4 5% } {% 25\left(\frac 4 {25}\right) - 8 + 4 } \\[6pt] % & = \frac{% 3\left(-\frac 8 {25}\right)+\frac 4 {25}+\frac{20}{25}% } {% 4 - 8 + 4 }\\[6pt] % & = \frac 0 0 \end{align*} $$
Find and divide out any common factors.
$$ \begin{align*} \lim_{x\to -\frac 2 5}\,% \frac{15x^3+x^2-2x} {25x^2 + 20x + 4}% % & = \lim_{x\to -\frac 2 5}\,% \frac{% x\blue{(5x+2)}(3x-1)% } {% \blue{(5x+2)}(5x+2)% } \\[6pt] % & = \lim_{x\to -\frac 2 5}\,% \frac{% x(3x-1)% } {% 5x+2% } \end{align*} $$
Evaluate the simpler limit .
$$ \begin{align*}% \lim_{x\to -\frac 2 5}\,% \frac{ x(3x-1)} {5x+2}% % & = \frac{% -\frac 2 5\left(% 3\left(-\frac 2 5\right) - 1% \right) } {% 5\left(-\frac 2 5\right) + 2 } \\[6pt] % & = \frac{% -\frac 2 5\left(% -\frac 6 5 - \frac 5 5% \right) } {% -2 + 2 } \\[6pt] % & = \frac{22/25} 0 \end{align*} $$.
$$ \displaystyle \lim_{x\to -\frac 2 5}\,% \frac{15x^3+x^2-2x} {25x^2 + 20x + 4} $$ does not exist.
