PEMFAKTORAN
\(\displaystyle \lim_{x \rightarrow a} \dfrac {1}{f(x)} - \dfrac {1}{g(x)} = \dfrac {1}{f(a)} - \dfrac {1}{g(a)} = \dfrac 10 - \dfrac 10 = \: \sim - \sim\)
Hasil \(\: \sim - \sim\) dapat diselesaikan dengan cara memfaktorkan \(f(x)\) dan \(g(x)\), dimana faktornya adalah \(x - a\)
\(\displaystyle \lim_{x \rightarrow a} \dfrac {1}{(x - a) \:.\: F(x)} - \dfrac {1}{(x - a) \:.\: G(x)} = \dfrac {G(x) - F(x)}{(x - a) \:.\: F(x) \:.\: G(x)}\)
Contoh
\(\displaystyle \lim_{x \rightarrow -3} \left(\frac{6}{x^2-9}-\frac{1}{x^2+5x+6}\right)\)
Jika \(x = -3\) disubstitusikan langsung, akan menghasilkan:
\(\displaystyle \lim_{x \rightarrow -3} \: \left(\frac{6}{x^2-9}-\frac{1}{x^2+5x+6}\right) = \frac{6}{(-3)^2-9}-\frac{1}{(-3)^2+5(-3)+6} = \frac 60 - \frac 10 = \: \sim - \sim\)
Untuk limit x mendekati −3, faktornya adalah \(x + 3\). Maka bentuk di atas dapat diubah menjadi:
\begin{equation*}
\begin{split}
& \lim_{x \rightarrow -3} \: \left(\frac{6}{x^2-9}-\frac{1}{x^2+5x+6} \right) \\\\
& \lim_{x \rightarrow -3} \: \left(\frac{6}{(x + 3)(x - 3)}-\frac{1}{(x + 3)(x + 2)} \right) \\\\
& \lim_{x \rightarrow -3} \: \frac{6(x + 2) - (x - 3)}{(x + 3)(x - 3)(x + 2)} \\\\
& \lim_{x \rightarrow -3} \: \frac{6x + 12 - x + 3}{(x + 3)(x - 3)(x + 2)} \\\\
& \lim_{x \rightarrow -3} \: \frac{5x + 15}{(x + 3)(x - 3)(x + 2)} \\\\
& \lim_{x \rightarrow -3} \: \frac{5(x + 3)}{(x + 3)(x - 3)(x + 2)} \\\\
& \lim_{x \rightarrow -3} \: \frac{5 \cancel {(x + 3)}}{\cancel {(x + 3)}(x - 3)(x + 2)} \\\\
& \lim_{x \rightarrow -3} \: \frac{5}{(x - 3)(x + 2)} \\\\
& \frac{5}{(-3 - 3)(-3 + 2)} \\\\
& \frac {5}{-6 \:.\: -1} \\\\
& \frac{5}{6}
\end{split}
\end{equation*}