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

Aug 5, 2014

Start by factoring the numerator:

$= {\lim}_{x \to 2} \left(\frac{\left(x + 3\right) \left(x - 2\right)}{x - 2}\right)$

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

$= {\lim}_{x \to 2} \left(x + 3\right)$

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 $\left(x - 2\right)$ 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.