Question

How do you find the limit `lim_(x->2)(x^2+x-6)/(x-2)` ?

Answer 1

Start by factoring the numerator:

`= lim_(x->2) (((x + 3)(x-2))/(x-2))`

We can see that the `(x - 2)` term will cancel off. Therefore, this limit is equivalent to:

`= lim_(x->2) (x + 3)`

It should now be easy to see what the limit evaluates to:

`= 5`

Let's take a look at a graph of what this function would look like, to see if our answer agrees:

The "hole" at `x = 2` is due to the `(x - 2)` term in the denominator. When `x = 2`, this term becomes `0`, and a division by zero occurs, resulting in the function being undefined at `x = 2`. However, the function is well-defined everywhere else, even when it gets extremely close to `x = 2`.

And, when `x` gets extremely close to `2`, `y` gets extremely close to `5`. This verifies what we demonstrated algebraically.