Hello, gmontalv!
\(\displaystyle \text{If }\,k(x)\:=\:\L\frac{x^3\,-\,2x\,-\,4}{x\,-\,2}\)
\(\displaystyle \;\;\;\text{find: }\,\lim_{x\to2}\,k(x)\)
Note that: \(\displaystyle \,k(2)\,=\,\frac{0}{0}\), an indetermante form.
This means that some cancelling is possible.
\(\displaystyle \;\;\) and we suspect that \(\displaystyle x\,-\,2\) is a factor of the numerator.
Using long (or synthetic) division, we factor the numerator.
\(\displaystyle \;\;\lim_{x\to2}\,\frac{(x\,-\,2)(x^2\,+\,2x\,+\,2)}{x\,-\,2} \;= \;\lim_{x\to2}\,(x^2\,+\,2x\,+\,2) \;= \;2^2\,+\,2\cdot2\,+\,2\;=\;10\)
