2.8: Algebraic Limits

A lot of times you don’t need to look at a graph or make a table to find a limit. Consider the following:

Find \(\lim_{x \to 3} 2x + 1\).

For this problem, we could make a table, or look at a graph, but \(2x + 1\) is such a nice function that there really isn’t any point in doing all that. Everything is happy at \(x = 3\), so we can just plug in that value.

\[\begin{align*} \lim_{x \to 3} 2x + 1 = 2(3) + 1 = 7. \end{align*}\]

In fact, any polynomial, logarithmic, or exponential function that you might run into is what is called continuous. Continuous means that the limit is what you get if you just plug the value into the function. So why bother with limits? Well, there are some functions that are not continuous, such as rational functions. And rational functions are exactly what crop up when taking derivatives. Consider this example:

Find \(\lim_{x \to 3} \frac{x^2 - 2x - 3}{x - 3}\).

Look at what happens when you plug in \(x = 3\).

\[\begin{align*} \lim_{x \to 3} \frac{x^2 - 2x - 3}{x - 3} & =^? \frac{(3)^2 - 2(3) - 3}{(3) - 3} \\ & = \frac{0}{0}. \end{align*}\]

This is the same problem we saw in the numerical limits section. The fraction \(\frac{0}{0}\) is called the indeterminate case. Here is where a graph or a table might be useful.

x	f(x)
2.5	3.5
2.9	3.9
2.99	3.99
2.999	3.999
2.9999	3.9999
3.0001	4.0001
3.001	4.001
3.01	4.01
3.1	4.1
3.5	4.5

Looks like the limit should be equal to \(4\).

But there is algebraic way. Just remember this: factor the top, and cancel like terms. Let’s see it in action.

\[\begin{align*} \lim_{x \to 3} \frac{x^2 - 2x - 3}{x - 3} &= \lim_{x \to 3} \frac{(x - 3)(x + 1)}{x - 3} \\ & = \lim_{x \to 3} \frac{\cancel{(x - 3)}(x+1)}{\cancel{x-3}} \\ & = \lim_{x \to 3} x + 1 \\ & = (3) + 1 \\ & = 4 \end{align*}\]

Notice though that it’s not right to say \(\frac{(x - 3)(x + 1)}{x - 3} = x+1\) always, since they are not equal when \(x = 3\). At \(x= 3\),\(\frac{(x - 3)(x + 1)}{x - 3}\) is not defined, and \(x + 1\) is. However, if you’re inside a limit as \(x \to 3\), then \(x\) is not equal to \(3\), it is just approaching \(3\). Hence it is okay to cancel those terms.

When dealing with rational expressions, sometimes you can’t cancel terms but can still find the value of the limit. For example,

Find \(\lim_{x \to -2} \frac{x^2 + 2}{x - 2}\),

The numerator does not factor. So there isn’t anything we can cancel. But if we just plug in \(x = -2\), we see

\[\begin{align*} \lim_{x \to -2} \frac{x^2 + 2}{x-2} & = \frac{(-2)^2 + 2}{(-2) - 2} \\ & = \frac{6}{-4} = - \frac{3}{2}. \end{align*}\]

In this example, the top is approaching \(6\), and the bottom \(-4\). Since there is no division by zero here, the limit value is just \(-\frac{3}{2}\). Easy peasy!

If you plug in a value to take a limit and you get a nonzero number divided by zero, you can just say the limit does not exist. For example,

\(\lim_{x \to -3} \frac{x}{x + 3},\)

we see

\[\begin{align*} \lim_{x \to -3} \frac{x}{x + 3} & = \frac{-3}{(-3) + 3} \\ & = \frac{-3}{0}. \end{align*}\]

Here, the denominator is approaching zero, while the numerator is holding steady at \(-3\). The result of this is dividing by smaller and smaller numbers, which means the value is getting bigger and bigger to infinity. Thus, the limit does not exist. Check out the table for this problem.

\(x\)	\(f(x)\)	Simplified
\(-3.5\)	\(\frac{-3.5}{-0.5}\)	\( = 7\)
\(-3.1\)	\(\frac{-3.1}{-0.1}\)	\( = 31\)
\(-3.01\)	\(\frac{-3.01}{-0.01}\)	\( = 301\)
\(-3.001\)	\(\frac{-3.001}{-0.001}\)	\( = 3001\)
\(-2.999\)	\(\frac{-2.999}{0.001}\)	\( = -2999\)
\(-2.99\)	\(\frac{-2.99}{0.01}\)	\( = -299\)
\(-2.9\)	\(\frac{-2.9}{0.1}\)	\( = -29\)

As you can see, \(f(x)\) is really large when \(x\) is close to \(-3\). In short: \(\frac{0}{0}\) is the indeterminate case where factoring and cancelling is a good idea. Anything else divided by zero is easy to determine: does not exist.

Algebraic limits

Find the following limit values algebraically in each case.

1. \(\lim_{x \to 4} \frac{7x - 4}{2x - 4}\)

   Here, we just need to plug in \(x = 4\) and we have \(\frac{7(4) - 4}{2(4) - 4} = \frac{24}{4} = \boxed{6}\). Since we didn’t divide by zero, this is continuous at this point, so we can just plug in the value.

2. \(\lim_{x \to 2} \frac{x^2 - 4}{x -2}\)

   Classic example where we factor the top and cancel.

   \(\lim_{x \to 2} \frac{x^2 - 4}{x -2} = \lim_{x \to 2} \frac{(x + 2)(x-2)}{x-2} = \lim_{x \to 2} x + 2 = \boxed{4}\)

3. \(\lim_{x \to 3} \frac{x + 3}{x - 3}\)

   In this example, we plug in \(x = 3\) and have \(\frac{6}{0}\). Since we are dividing by zero, this \(\boxed{\text{limit does not exist}}\).

4. \(\lim_{x \to -1} \frac{x^2 + 6x + 5}{x +1}\)

   Another example of factoring and cancelling.

   \(\lim_{x \to -1} \frac{x^2+6x + 5}{x +1} = \lim_{x \to -1} \frac{(x + 1)(x + 5)}{x+1} = \lim_{x \to -1} x + 5 = \boxed{4}\)

5. \(\lim_{x \to 1} \frac{(x - 1)^3}{x - 1}\)

   Another example of factoring and cancelling, but this time the top is already factored!

   \(\lim_{x \to 1} \frac{(x-1)^3}{x-1} = (x-1)^2 = (1-1)^2 = 0^2 = \boxed{0}.\)