Properties of the Limits
Related calculator: Limit Calculator
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.