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.
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.
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.
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.
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)
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
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