Most of the time, it's more precise to find the limit using algebra.

When finding a limit of the form  , 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.

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.

x²-x-2 = (x + 1)(x-2)

and

x²-5x + 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 quotient f(x) can be factored as

 which equals (x-1) for every value of x except -1 (when x = -1, the quotient 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

by instead finding

.

Practice: