# Evaluating limits

## Simplifying

Limits obey a set of laws, and the basic ones are worth remembering. First, the limit of a sum is equal to the sum of the limits:

$limx→a⁡(f(x)+g(x))=limx→a⁡f(x)+limx→a⁡g(x).$

The same thing goes for subtraction, multiplication, and division.

## Direct substitution

If a function $f$ is continuous at $a$ and $f\left(a\right)$ is defined, then

$limx→a⁡f(x)=f(a).$

You’ll learn about continuity in the next section, but for now you can apply this rule to polynomials, rationals, exponentials, logarithms, and trigonometric functions—everything you saw in Advanced Functions.

For example, we can use direct substitution in

$limx→3⁡x2=32=9.$

## Factoring

Sometimes attempting direct substitution will give us zero over zero—we call this result indeterminate (this is distinct from undefined). If the expression looks like it can be factored, try doing so. Sometimes this will result in a factor dividing out, and then you can proceed with direct substitution after that.

For example, we can factor the numerator to evaluate this limit:

$limx→0⁡7x−x2x=limx→0⁡x(7−x)x=limx→0⁡7−x=7−0=7.$

## Rationalizing

If factoring doesn’t work, we can try rationalizing to get around the indeterminate case. This generally works if you have a binomial with one or two radicals in it. To do this, we multiply by the conjugate of binomial over itself and then use the difference of squares.

An expression of the form ${a}^{2}-{b}^{2}$ can be factored to give $\left(a+b\right)\left(a-b\right)\text{.}$ In this expression, $a+b$ is the conjugate of $a-b\text{,}$ and vice versa.

For example, multiplying the following by the conjugate over itself gives us

$limx→0⁡x+1−1x=limx→0⁡(x+1−1x)(x+1+1x+1+1),$

which simplifies to give the value of the limit,

$limx→0⁡(x+1)−1(x(x+1+1))=limx→0⁡1x+1+1=10+1+1=12.$

## Change of variable

Yet another strategy to try when tackling the indeterminate is to change the variable. It’s hard to explain this method in words, so I’ll just demonstrate it with an example.

Suppose we want to evaluate the limit

$limx→0⁡(2x+1)13−1x.$

Let’s change the variable. In general, when we have a fractional exponent as we do here, we want to get rid of it. To do this, we will let ${u}^{3}=2x+1\text{.}$ This means that

$(2x+1)13=u$ and $x=12(u3−1).$

Now we can change the variable in the limit expression:

$limx→0⁡u−112(u3−1)=limx→0⁡u−112(u−1)(u2+u+1)=limx→0⁡2u2+u+1.$

We were able to divide out a factor, which is exactly what we wanted. Now we can back-substitute our original variable and evaluate:

$limx→0⁡2(2x+1)23+(2x+1)13+1=2123+113+1=23.$

## One-sided limit

Some limits cannot be evaluated because they do not exist. To prove this, we have to show that the limit from the left does not equal the limit from the right. Consider the following limit:

$limx→4⁡|x−4|x−4.$

We get an indeterminate answer if we try direct substitution. As you might have guessed, we need to evaluate the limit from both sides. Graphing the function turns out to be very helpful:

This graph makes it obvious that

$limx→4−⁡|x−4|x−4=−1$ and $limx→4+⁡|x−4|x−4=1,$

and since these are not equal, the limit does not exist.