# One-Sided Limits

Now we can extend concept of limit.

Definition. We write that $\lim_{{{x}\to{{a}}^{{-}}}}{f{{\left({x}\right)}}}={L}$ and say "the limit of f(x), as x approaches a from the left, equals L" if for any $\epsilon>{0}$ there exists $\delta>{0}$ such that ${\left|{f{{\left({x}\right)}}}-{L}\right|}<\epsilon$ when ${\left|{x}-{a}\right|}<\delta$ and ${x}>{a}$.

In other words we are interested in behavior of function near ${a}$ and only to the left from ${a}$.

Definition. We write that $\lim_{{{x}\to{{a}}^{+}}}{f{{\left({x}\right)}}}={L}$ and say "the limit of f(x), as x approaches a from the right, equals L" if for any $\epsilon>{0}$ there exists $\delta>{0}$ such that ${\left|{f{{\left({x}\right)}}}-{L}\right|}<\epsilon$ when ${\left|{x}-{a}\right|}<\delta$ and ${x}<{a}$.

In other words we are interested in behavior of function near ${a}$ and only to the right from ${a}$.

One-sided limits often arise for piecewise defined functions.

Example 1. Find $\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}}$, $\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}$, $\lim_{{{x}\to{1}}}{f{{\left({x}\right)}}}$ where ${f{{\left({x}\right)}}}={\left\{\begin{array}{c}{2}{\quad\text{if}\quad}{x}\ge{1}\\{1}{\quad\text{if}\quad}{x}<{1}\\ \end{array}\right.}$.

As ${x}$ approaches 1 from the left ${f{{\left({x}\right)}}}$ approaches 1, thus $\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}={1}$.

As ${x}$ approaches 1 from the right ${f{{\left({x}\right)}}}$ approaches 2, thus $\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}}={2}$.

So, there is no single number that ${f{{\left({x}\right)}}}$ approaches as ${x}$ approaches 1, thus $\lim_{{{x}\to{1}}}{f{{\left({x}\right)}}}$ doesn't exist.

Fact. $\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={L}$ if and only if $\lim_{{{x}\to{{a}}^{{-}}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{a}}^{+}}}{f{{\left({x}\right)}}}={L}$.

This fact is quite useful because often it is easier to find one-sided limits and to check whether they are equal to make conclusion about limit.

Example 2. The graph of the function ${f{{\left({x}\right)}}}$ is shown. Find (if they exist) values of the following limits:

1. $\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}$
2. $\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}}$
3. $\lim_{{{x}\to{1}}}{f{{\left({x}\right)}}}$
4. $\lim_{{{x}\to{{4}}^{{-}}}}{f{{\left({x}\right)}}}$
5. $\lim_{{{x}\to{4}+}}{f{{\left({x}\right)}}}$
6. $\lim_{{{x}\to{4}}}{f{{\left({x}\right)}}}$

From graph it is seen that ${f{{\left({x}\right)}}}$ approaches 2 as ${x}$ approaches 1 from the left and ${f{{\left({x}\right)}}}$ approaches 3 when ${x}$ approaches 1 from the right.

Therefore, $\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}={2}$ and $\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}}={3}$.

Since $\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}\ne\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}}$ then $\lim_{{{x}\to{1}}}{f{{\left({x}\right)}}}$ doesn't exist.

Also from graph it can be seen that ${f{{\left({x}\right)}}}$ approaches 1 as ${x}$ approaches 4 from the left and ${f{{\left({x}\right)}}}$ approaches 1 when ${x}$ approaches 4 from the right. Therefore, $\lim_{{{x}\to{{4}}^{{-}}}}{f{{\left({x}\right)}}}={1}$ and $\lim_{{{x}\to{{4}}^{+}}}{f{{\left({x}\right)}}}={1}$.

Since $\lim_{{{x}\to{{4}}^{{-}}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{4}}^{+}}}{f{{\left({x}\right)}}}={1}$ then $\lim_{{{x}\to{4}}}{f{{\left({x}\right)}}}={1}$.

Despite this fact notice that ${f{{\left({4}\right)}}}\ne{1}$.

Example 3. Find $\lim_{{{x}\to{0}}}{\left|{x}\right|}$.

Recall that ${\left|{x}\right|}={\left\{\begin{array}{c}{x}{\quad\text{if}\quad}{x}\ge{0}\\-{x}{\quad\text{if}\quad}{x}<{0}\\ \end{array}\right.}$.

Therefore, for ${x}>{0}$ ${\left|{x}\right|}={x}$, thus $\lim_{{{x}\to{0}+}}{\left|{x}\right|}=\lim_{{{x}\to{{0}}^{+}}}{x}={0}$.

Since for ${x}<{0}$ ${\left|{x}\right|}=-{x}$, then $\lim_{{{x}\to{{0}}^{{-}}}}{\left|{x}\right|}=\lim_{{{x}\to{{0}}^{{-}}}}-{x}={0}$.

We see that $\lim_{{{x}\to{{0}}^{+}}}{\left|{x}\right|}=\lim_{{{x}\to{{0}}^{{-}}}}{\left|{x}\right|}={0}$. That's why $\lim_{{{x}\to{0}}}{\left|{x}\right|}={0}$.

Example 4. Find $\lim_{{{x}\to{0}}}\frac{{{\left|{x}\right|}}}{{x}}$.

For ${x}>{0}$ ${\left|{x}\right|}={x}$, thus $\lim_{{{x}\to{0}+}}\frac{{{\left|{x}\right|}}}{{x}}=\lim_{{{x}\to{{0}}^{+}}}\frac{{x}}{{x}}=\lim_{{{x}\to{{0}}^{+}}}{1}={1}$.

For ${x}<{0}$ ${\left|{x}\right|}=-{x}$, thus $\lim_{{{x}\to{{0}}^{{-}}}}\frac{{{\left|{x}\right|}}}{{x}}=\lim_{{{x}\to{{0}}^{{-}}}}\frac{{-{x}}}{{x}}=\lim_{{{x}\to{{0}}^{{-}}}}-{1}=-{1}$.

We see that $\lim_{{{x}\to{{0}}^{+}}}\frac{{{\left|{x}\right|}}}{{x}}\ne\lim_{{{x}\to{{0}}^{{-}}}}\frac{{{\left|{x}\right|}}}{{x}}$. That's why $\lim_{{{x}\to{0}}}\frac{{{\left|{x}\right|}}}{{x}}$ doesn't exist.