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.