How would you "remove the discontinuity" of $f$ ? In other words, how would you define $f(4)$ in order to make $f$ continuous at $x=4$?
$$f(x) = \dfrac{x^2-x-12}{x-4}$$
5 Answers
You have $f(x)= \dfrac{x^2-x-12}{x-4}$. Notice that $x=4$ is not in the domain of the function since then you would be dividing by $0$. However, if $x \neq 4$, then we have $$ \require{cancel} f(x)= \dfrac{x^2-x-12}{x-4}= \dfrac{(x-4)(x+3)}{x-4}=\dfrac{\cancel{(x-4)}(x+3)}{\cancel{x-4}}= x+ 3 $$ Notice that $x+3$ gets 'close' to $4+3=7$ when $x$ is 'close' to $x=4$. Then if we want to define a function which is equal to $f(x)$ when $x \neq 4$, is defined at $x=4$, and is continuous, we have to choose $$ g(x)= \begin{cases} \dfrac{x^2-x-12}{x-4}= x+3, & x\neq 4 \\ 7, & x=4 \end{cases} $$
Using the fact that $x^2-x-12=(x-4)(x+3)$.
Note that $x^2-x-12=(x-4)(x+3)$.
Now, $$\lim\limits_{x\to 4} f(x)=\lim_{x\to 4} x+3=7$$.
So, define $f(4)=7$.
$f(x) = \frac{x^2-x-12}{x-4}$
$f(x) = \frac{(x-4)(x+3)}{(x-4)}$
$\lim_{x\to4}f(x) = \frac{x^2-x-12}{x-4} = \lim_{x\to4}\frac{(x-4)(x+3)}{(x-4)} = 7$
So redefine $f(x)$ as ;
$f(x) = \begin{cases}\frac{x^2-x-12}{x-4}&x\ne4\\7&x= 4\end{cases}$
if $ x \neq 4 $
$f(x)= \dfrac{x^2-x-12}{x-4}= \dfrac{(x-4)(x+3)}{x-4}=\dfrac{1(x+3)}{1}= x+ 3$
so we have f(x)=x+3 (for $ x \neq 4$)
$\lim_{x\to4}f(x) = \lim_{x\to4}(x+3) $= 7
we want to make f continuous for this reason the ammount of function at 4 should be equall to $$\lim_{x\to4}f(x)=7$$
so we define a function that in x=4 it equall to $\lim_{x\to4}f(x)=7$
$$g(x)= \begin{cases} \dfrac{x^2-x-12}{x-4}= x+3, & x\neq 4 \\ 7, & x=4 \end{cases}$$
