Properties of the Limits

Now it is time to give the properties of the limits that will allow to calculate the limits easier.

Suppose that $c$ is a constant and the limits $\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}$ and $\lim_{{{x}\to{a}}}{g{{\left({x}\right)}}}$ exist; then, the following laws hold:

Law 1. $\lim_{{{x}\to{a}}}{c}={c}$ .

Law 2. $\lim_{{{x}\to{a}}}{\left({f{{\left({x}\right)}}}\pm{g{{\left({x}\right)}}}\right)}=\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}\pm\lim_{{{x}\to{a}}}{g{{\left({x}\right)}}}$ . This is true for any number of functions.

Law 3. $\lim_{{{x}\to{a}}}{\left({c}{f{{\left({x}\right)}}}\right)}={c}\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}$ .

Law 4. $\lim_{{{x}\to{a}}}{\left({f{{\left({x}\right)}}}{g{{\left({x}\right)}}}\right)}=\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}\cdot\lim_{{{x}\to{a}}}{g{{\left({x}\right)}}}$ .

Law 5. $\lim_{{{x}\to{a}}}{\left(\frac{{f{{\left({x}\right)}}}}{{g{{\left({x}\right)}}}}\right)}=\frac{{\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}}}{{\lim_{{{x}\to{a}}}{g{{\left({x}\right)}}}}}$ provided that $\lim_{{{x}\to{a}}}{g{{\left({x}\right)}}}\ne{0}$ .

Law 6. $\lim_{{{x}\to{a}}}{{\left({f{{\left({x}\right)}}}\right)}}^{{n}}={{\left(\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}\right)}}^{{n}}$ , where ${n}$ is a real number.

Law 7. $\lim_{{{x}\to{a}}}{\sqrt[{{n}}]{{{f{{\left({x}\right)}}}}}}={\sqrt[{{n}}]{{\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}}}}$ .

Law 8. $\lim_{{{x}\to{a}}}{x}={a}$ .

Law 9. $\lim_{{{x}\to{a}}}{{x}}^{{n}}={{a}}^{{n}}$ .

These laws are also true for one-sided limits.

Now, let's go through a couple of examples.

Example 1. Find $\lim_{{{x}\to{2}}}{\left({{x}}^{{3}}-{3}{{x}}^{{2}}+{4}{x}+{1}\right)}$ .

$\lim_{{{x}\to{2}}}{\left({{x}}^{{3}}-{3}{{x}}^{{2}}+{4}{x}+{1}\right)}=\lim_{{{x}\to{2}}}{\left({{x}}^{{3}}\right)}-\lim_{{{x}\to{2}}}{\left({3}{{x}}^{{2}}\right)}+\lim_{{{x}\to{2}}}{\left({4}{x}\right)}+\lim_{{{x}\to{2}}}{\left({1}\right)}=$ according to law 2

$=\lim_{{{x}\to{2}}}{\left({{x}}^{{3}}\right)}-{3}\lim_{{{x}\to{2}}}{\left({{x}}^{{2}}\right)}+{4}\lim_{{{x}\to{2}}}{\left({x}\right)}+\lim_{{{x}\to{2}}}{\left({1}\right)}=$ according to law 3

$={{2}}^{{3}}-{3}\cdot{{2}}^{{2}}+{4}\cdot{2}+{1}={5}$ . according to laws 1, 6, and 8.

Let's do another example.

Example 2. Find $\lim_{{{x}\to{1}}}{\left(\frac{{{{x}}^{{2}}+{3}{x}}}{{{1}-{2}{x}}}\right)}$ .

We need to use law 5, but first we need to make sure that $\lim_{{{x}\to{1}}}{\left({1}-{2}{x}\right)}\ne{0}$ .

$\lim_{{{x}\to{1}}}{\left({1}-{2}{x}\right)}=\lim_{{{x}\to{1}}}{\left({1}\right)}-\lim_{{{x}\to{1}}}{\left({2}{x}\right)}=$ according to law 2

$=\lim_{{{x}\to{1}}}{\left({1}\right)}-{2}\lim_{{{x}\to{1}}}{\left({x}\right)}$ according to law 3

$={1}-{2}\cdot{1}=-{1}\ne{0}$ . according to laws 1 and 8.

So, we can use law 5:

$\lim_{{{x}\to{1}}}{\left(\frac{{{{x}}^{{2}}+{3}{x}}}{{{1}-{2}{x}}}\right)}=\frac{{\lim_{{{x}\to{1}}}{\left({{x}}^{{2}}+{3}{x}\right)}}}{{\lim_{{{x}\to{1}}}{\left({1}-{2}{x}\right)}}}=\frac{{\lim_{{{x}\to{1}}}{\left({{x}}^{{2}}+{3}{x}\right)}}}{{-{1}}}=-{\left(\lim_{{{x}\to{1}}}{\left({{x}}^{{2}}+{3}{x}\right)}\right)}=$

$=-{\left(\lim_{{{x}\to{1}}}{\left({{x}}^{{2}}\right)}+\lim_{{{x}\to{1}}}{\left({3}{x}\right)}\right)}=$ according to law 2

$=-{\left(\lim_{{{x}\to{1}}}{\left({{x}}^{{2}}\right)}+{3}\lim_{{{x}\to{1}}}{\left({x}\right)}\right)}=$ according to law 1

$=-{\left({{1}}^{{2}}+{3}\cdot{1}\right)}=-{4}$ . according to laws 6 and 8.

Note that in the above examples we can just use direct substitution (in example 1, we could plug $2$ , and in example 2, we could put $1$ ) to obtain the same correct answer.

In example 1, $\lim_{{{x}\to{2}}}{\left({{x}}^{{3}}-{3}{{x}}^{{2}}+{4}{x}+{1}\right)}={{2}}^{{3}}-{3}\cdot{{2}}^{{2}}+{4}\cdot{2}+{1}={5}$ , and in example 2, $\lim_{{{x}\to{1}}}{\left(\frac{{{{x}}^{{2}}+{3}{x}}}{{{1}-{2}{x}}}\right)}=\frac{{{{1}}^{{2}}+{3}\cdot{1}}}{{{1}-{2}\cdot{1}}}=-{4}$ .

Example 1 was to find the limit of a polynomial, and example 2 was to find the limit of a rational function. In fact, for these types of functions, we can always use direct substituion.

Direct Substitution Property. If ${f{}}$ is a polynomial or a rational function and ${a}$ is in the domain of ${f{}}$ , this means that $\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={f{{\left({a}\right)}}}$ .

Example 3. Find $\lim_{{{x}\to{1}}}\frac{{{{x}}^{{3}}-{1}}}{{{x}-{1}}}$ .

We can't use law 5 because $\lim_{{{x}\to{1}}}{\left({x}-{1}\right)}={0}$ ; we can't use the direct subsitution property either because ${x}={1}$ is not in the domain.

However, we can use algebra to simplify the function. Note that ${{x}}^{{3}}-{1}={\left({x}-{1}\right)}{\left({{x}}^{{2}}+{x}+{1}\right)}$ .

$\lim_{{{x}\to{1}}}{\left(\frac{{{{x}}^{{3}}-{1}}}{{{x}-{1}}}\right)}=\lim_{{{x}\to{1}}}{\left(\frac{{{\left({x}-{1}\right)}{\left({{x}}^{{2}}+{x}+{1}\right)}}}{{{x}-{1}}}\right)}=$

On this step, we can cancel out the common term in the numerator and the denominator because ${x}$ approaches $1$ but doesn't equal $1$ (recall from the definition of the limit that we are interested in the behavior of the function near the point of approach), and thus ${x}-{1}\ne{0}$ .

$=\lim_{{{x}\to{1}}}{\left({{x}}^{{2}}+{x}+{1}\right)}={{1}}^{{2}}+{1}+{1}={3}$ .

Let's proceed to another example.

Example 4. Find $\lim_{{{x}\to{0}}}{g{{\left({x}\right)}}}$ , where ${g{{\left({x}\right)}}}={\left\{\begin{array}{c}{{x}}^{{2}}+{3}{\quad\text{if}\quad}{x}\ne{0}\\{0}{\quad\text{if}\quad}{x}={0}\\ \end{array}\right.}$ .

Here, ${g{}}$ is defined at $0$ , and ${g{{\left({0}\right)}}}={0}$ , but the limit doesn't depend on the value at the point $0$ . We investigate the behavior near $0$ , and near $0$ the function is defined as ${{x}}^{{2}}+{3}$ ; thus, $\lim_{{{x}\to{0}}}{g{{\left({x}\right)}}}=\lim_{{{x}\to{0}}}{\left({{x}}^{{2}}+{3}\right)}={{0}}^{{2}}+{3}={3}$ .

This one was quick, so let's move on to the last example.

Example 5. Find $\lim_{{{x}\to{0}}}\frac{{\sqrt{{{{x}}^{{2}}+{4}}}-{2}}}{{{{x}}^{{2}}}}$ .

We can't apply the quotient law (law 5) because the limit of the denominator equals $0$ .

So, we need to perform some algebraic manipulations.

Let's rationalize the numerator:

$\lim_{{{x}\to{0}}}\frac{{\sqrt{{{{x}}^{{2}}+{4}}}-{2}}}{{{{x}}^{{2}}}}=\lim_{{{x}\to{0}}}{\left(\frac{{\sqrt{{{{x}}^{{2}}+{4}}}-{2}}}{{{{x}}^{{2}}}}\ \frac{{\sqrt{{{{x}}^{{2}}+{4}}}+{2}}}{{\sqrt{{{{x}}^{{2}}+{4}}}+{2}}}\right)}=\lim_{{{x}\to{0}}}\frac{{{\left({{x}}^{{2}}+{4}\right)}-{{2}}^{{2}}}}{{{{x}}^{{2}}{\left(\sqrt{{{{x}}^{{2}}+{4}}}+{2}\right)}}}=$

$=\lim_{{{x}\to{0}}}\frac{{{{x}}^{{2}}}}{{{{x}}^{{2}}{\left(\sqrt{{{{x}}^{{2}}+{4}}}+{2}\right)}}}=\lim_{{{x}\to{0}}}{\left(\frac{{1}}{{\sqrt{{{{x}}^{{2}}+{4}}}+{2}}}\right)}={\left(\frac{{1}}{{\sqrt{{\lim_{{{x}\to{0}}}{\left({{x}}^{{2}}+{4}\right)}}}+{2}}}\right)}=$

$=\frac{{1}}{{\sqrt{{{{0}}^{{2}}+{4}}}+{2}}}=\frac{{1}}{{4}}$ .

So, we often need to do some algebraic manipulations before evaluating the limit.