One-Sided Limits

Now we can extend concept of limit.

Definition. We write that limxaf(x)=L\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 ϵ>0\epsilon>{0} there exists δ>0\delta>{0} such that f(x)L<ϵ{\left|{f{{\left({x}\right)}}}-{L}\right|}<\epsilon when xa<δ{\left|{x}-{a}\right|}<\delta and x>a{x}>{a}.

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

Definition. We write that limxa+f(x)=L\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 ϵ>0\epsilon>{0} there exists δ>0\delta>{0} such that f(x)L<ϵ{\left|{f{{\left({x}\right)}}}-{L}\right|}<\epsilon when xa<δ{\left|{x}-{a}\right|}<\delta and x<a{x}<{a}.

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

One-sided limits often arise for piecewise defined functions.

Example 1. Find limx1+f(x)\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}}, limx1f(x)\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}, limx1f(x)\lim_{{{x}\to{1}}}{f{{\left({x}\right)}}} where f(x)={2ifx11ifx<1{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.}.one-sided limit

As x{x} approaches 1 from the left f(x){f{{\left({x}\right)}}} approaches 1, thus limx1f(x)=1\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}={1}.

As x{x} approaches 1 from the right f(x){f{{\left({x}\right)}}} approaches 2, thus limx1+f(x)=2\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}}={2}.

So, there is no single number that f(x){f{{\left({x}\right)}}} approaches as x{x} approaches 1, thus limx1f(x)\lim_{{{x}\to{1}}}{f{{\left({x}\right)}}} doesn't exist.

Fact. limxaf(x)=L\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={L} if and only if limxaf(x)=limxa+f(x)=L\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(x){f{{\left({x}\right)}}} is shown. Find (if they exist) values of the following limits:

one sided limit example

  1. limx1f(x)\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}
  2. limx1+f(x)\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}}
  3. limx1f(x)\lim_{{{x}\to{1}}}{f{{\left({x}\right)}}}
  4. limx4f(x)\lim_{{{x}\to{{4}}^{{-}}}}{f{{\left({x}\right)}}}
  5. limx4+f(x)\lim_{{{x}\to{4}+}}{f{{\left({x}\right)}}}
  6. limx4f(x)\lim_{{{x}\to{4}}}{f{{\left({x}\right)}}}

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

Therefore, limx1f(x)=2\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}={2} and limx1+f(x)=3\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}}={3}.

Since limx1f(x)limx1+f(x)\lim_{{{x}\to{{1}}^{{-}}}}{f{{\left({x}\right)}}}\ne\lim_{{{x}\to{{1}}^{+}}}{f{{\left({x}\right)}}} then limx1f(x)\lim_{{{x}\to{1}}}{f{{\left({x}\right)}}} doesn't exist.

Also from graph it can be seen that f(x){f{{\left({x}\right)}}} approaches 1 as x{x} approaches 4 from the left and f(x){f{{\left({x}\right)}}} approaches 1 when x{x} approaches 4 from the right. Therefore, limx4f(x)=1\lim_{{{x}\to{{4}}^{{-}}}}{f{{\left({x}\right)}}}={1} and limx4+f(x)=1\lim_{{{x}\to{{4}}^{+}}}{f{{\left({x}\right)}}}={1}.

Since limx4f(x)=limx4+f(x)=1\lim_{{{x}\to{{4}}^{{-}}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{{4}}^{+}}}{f{{\left({x}\right)}}}={1} then limx4f(x)=1\lim_{{{x}\to{4}}}{f{{\left({x}\right)}}}={1}.

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

Example 3. Find limx0x\lim_{{{x}\to{0}}}{\left|{x}\right|}.

Recall that x={xifx0xifx<0{\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{x}>{0} x=x{\left|{x}\right|}={x}, thus limx0+x=limx0+x=0\lim_{{{x}\to{0}+}}{\left|{x}\right|}=\lim_{{{x}\to{{0}}^{+}}}{x}={0}.

Since for x<0{x}<{0} x=x{\left|{x}\right|}=-{x}, then limx0x=limx0x=0\lim_{{{x}\to{{0}}^{{-}}}}{\left|{x}\right|}=\lim_{{{x}\to{{0}}^{{-}}}}-{x}={0}.

We see that limx0+x=limx0x=0\lim_{{{x}\to{{0}}^{+}}}{\left|{x}\right|}=\lim_{{{x}\to{{0}}^{{-}}}}{\left|{x}\right|}={0}. That's why limx0x=0\lim_{{{x}\to{0}}}{\left|{x}\right|}={0}.

Example 4. Find limx0xx\lim_{{{x}\to{0}}}\frac{{{\left|{x}\right|}}}{{x}}.

For x>0{x}>{0} x=x{\left|{x}\right|}={x}, thus limx0+xx=limx0+xx=limx0+1=1\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{x}<{0} x=x{\left|{x}\right|}=-{x}, thus limx0xx=limx0xx=limx01=1\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 limx0+xxlimx0xx\lim_{{{x}\to{{0}}^{+}}}\frac{{{\left|{x}\right|}}}{{x}}\ne\lim_{{{x}\to{{0}}^{{-}}}}\frac{{{\left|{x}\right|}}}{{x}}. That's why limx0xx\lim_{{{x}\to{0}}}\frac{{{\left|{x}\right|}}}{{x}} doesn't exist.