# One-Sided Limits

## Related Calculator: Limit Calculator

Now we can extend concept of limit.

**Definition**. We write that `lim_(x->a^-)f(x)=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 `|f(x)-L|<epsilon` when `|x-a|<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->a^+)f(x)=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 `|f(x)-L|<epsilon` when `|x-a|<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->1^+)f(x)`, `lim_(x->1^-)f(x)`, `lim_(x->1)f(x)` where `f(x)={(2 if x>=1),(1 if x<1):}`.

As `x` approaches 1 from the left `f(x)` approaches 1, thus `lim_(x->1^-)f(x)=1`.

As `x` approaches 1 from the right `f(x)` approaches 2, thus `lim_(x->1^+)f(x)=2`.

So, there is no single number that `f(x)` approaches as `x` approaches 1, thus `lim_(x->1)f(x)` doesn't exist.

**Fact**. `lim_(x->a)f(x)=L` if and only if `lim_(x->a^-)f(x)=lim_(x->a^+)f(x)=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(x)` is shown. Find (if they exist) values of the following limits:

- `lim_(x->1^-)f(x)`
- `lim_(x->1^+)f(x)`
- `lim_(x->1)f(x)`
- `lim_(x->4^-)f(x)`
- `lim_(x->4+)f(x)`
- `lim_(x->4)f(x)`

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

Therefore, `lim_(x->1^-)f(x)=2` and `lim_(x->1^+)f(x)=3`.

Since `lim_(x->1^-)f(x)!=lim_(x->1^+)f(x)` then `lim_(x->1)f(x)` doesn't exist.

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

Since `lim_(x->4^-)f(x)=lim_(x->4^+)f(x)=1` then `lim_(x->4)f(x)=1`.

Despite this fact notice that `f(4)!=1`.

**Example 3**. Find `lim_(x->0)|x|`.

Recall that `|x|={(x if x>=0),(-x if x<0):}`.

Therefore, for `x>0` `|x|=x`, thus `lim_(x->0+)|x|=lim_(x->0^+)x=0`.

Since for `x<0` `|x|=-x`, then `lim_(x->0^-)|x|=lim_(x->0^-)-x=0`.

We see that `lim_(x->0^+)|x|=lim_(x->0^-)|x|=0`. That's why `lim_(x->0)|x|=0`.

**Example 4**. Find `lim_(x->0)(|x|)/x`.

For `x>0` `|x|=x`, thus `lim_(x->0+)(|x|)/x=lim_(x->0^+)x/x=lim_(x->0^+)1=1`.

For `x<0` `|x|=-x`, thus `lim_(x->0^-)(|x|)/x=lim_(x->0^-)(-x)/x=lim_(x->0^-)-1=-1`.

We see that `lim_(x->0^+)(|x|)/x!=lim_(x->0^-)(|x|)/x`. That's why `lim_(x->0)(|x|)/x` doesn't exist.