Question

Is the function `(x^2-x-12)/(x-4)` continuous or not? Because I drew an online graph and it appeared continuous...?

Also, (in continuation) , why can we find the `lim x->4 (x^2-x-12)/(x-4)` only after simplifying the expression to (x+3)? Aren't the two equivalent? Can someone explain the thought process and intuition behind these calculus basics?

Answer 1

Below

Explanation:

First question:
The graph is discontinuous at the point `x=4`. When you look at the graph before it is simplified, you will notice that the denominator is equal to `x-4`. When you sub in the point `x=4`, your denominator is equal to 0. That means that your graph is undefined or no solution at that point, so technically, there should be an open circle at `x=4` on the graph.

Second question:
The whole point of limits is to see what happens to the graph when you are approaching a number. So when your graph is approaching 4 on both sides, you will notice that it converges to `y=7` when `x=4` or in other words, the point `(4,7)`. This does not mean that it is AT `(4,7)` but rather it is approaching `(4,7)`. So you basically have y values such as `y=6.99999999` or `y=7.000001111111` or values of y which are very very close to 7 but are not equal to 7.

Technically, your limit should have been written like this:
`y=lim_(x ->4)(x^2-x-12)/(x-4)`

graph{(y-x^2+x+12)(y-x+4)=0 [-10, 10, -5, 5]}

When looking at the graph above, you will notice that at `x=4`, the two graphs meet. At that point, you will realise that if you divide `x^2-x-12` by `x-4` when `x=4`, you will end up getting 0 which as you can see is not the case.