# Limits via Algebra

Most of the time, it's more precise (and a lot faster) to find limits using algebra.

When finding a limit of the form , where *f*(*x*) is just one nice algebraic expression, the first thing to do is plug *a* into the function to see if it exists. If *f*(*a*) exists, then that's the answer. This won't work for piecewise functions, since there could be gaps separating the pieces of the function.

### Sample Problem

Find .

The first thing we do is plug in 3 and see if we find a value.

That's a perfectly good fraction, therefore

Here's an example of how we might not find a solution.

### Sample Problem

Find .

The first thing we do is plug in 0 and see if we find a number.

= ?

The fraction is undefined, since we can't divide by zero.

### Sample Problem

Find .

The first thing we do is see if we can plug in 2.

This answer is undefined.

When we're asked to find the limit of a quotient, if we plug in a number and find there's a good chance we can do something about it: we simplify the quotient and try again.

### Sample Problem

Find .

We already tried to plug in 2 and that didn't work. We'll simplify the fraction by factoring the polynomials.

*x*^{2}* – x* – 2 = (*x* + 1)(*x* – 2)

and

*x*^{2} – 5*x* + 6 = (*x* – 2)(*x* – 3).

Now we can see why we got before: 2 is a root of the polynomial in the numerator, and also a root of the polynomial in the denominator. Here's the point where we can do something useful: cancel the term (*x* – 2) from the numerator and denominator:

.

Finally, put in 2 again:

We can now say

= 2.

Why does this work? Let

* *

Now we'll think about .

The numerator of *f*(*x*) can be factored as

which equals (*x* – 1) for every value of *x* except -1 (when *x* = -1, the function is undefined).

We'll say that again, because it's important. After factoring *f*, we see that we can think of it as

.

If we graph this, we find the line *x* – 1 with a "hole" in the graph at *x* = -1 since* f*(-1) is undefined:

It's like the function *f*(*x*) is trying to be *x* – 1, but failing at one spot (poor function!). Here's the good news: since *f*(*x*) is trying to be *x* – 1, we can find the limit by instead finding

*.*