Definition of Limit of a Function

Cauchy and Heine Definitions of Limit

Let f (x) be a function that is defined on an open interval X containing x = a. (The value f (a) need not be defined.)

The number L is called the limit of function f (x) as x → a if and only if, for every ε > 0 there exists δ > 0 such that

\[\left| {f\left( x \right) - L} \right| \lt \varepsilon ,\]

whenever

\[0 \lt \left| {x - a} \right| \lt \delta .\]

This definition is known as ε−δ- or Cauchy definition for limit.

There's also the Heine definition of the limit of a function, which states that a function \(f\left( x \right)\) has a limit \(L\) at \(x = a\), if for every sequence \(\left\{ {{x_n}} \right\}\) (with \(\left\{ {{x_n}} \right\}\) not equal to \(a\) for all \(n\) ), which has a limit at \(a,\) the sequence \(\left\{f\left( {{x_n}} \right)\right\}\) converges to \(L.\)

The Heine and Cauchy definitions of limit of a function are equivalent.

One-Sided Limits

Let \(\lim\limits_{x \to a - 0} \) denote the limit as \(x\) goes toward \(a\) by taking on values of \(x\) such that \(x \lt a\). The corresponding limit \(\lim\limits_{x \to a - 0} f\left( x \right)\) is called the left-hand limit of \(f\left( x \right)\) at the point \(x = a\).

Similarly, let \(\lim\limits_{x \to a + 0} \) denote the limit as \(x\) goes toward \(a\) by taking on values of \(x\) such that \(x \gt a\). The corresponding limit \(\lim\limits_{x \to a + 0} f\left( x \right)\) is called the right-hand limit of \(f\left( x \right)\) at \(x = a\).

Note that the \(2\)-sided limit \(\lim\limits_{x \to a} f\left( x \right)\) exists only if both one-sided limits exist and are equal to each other, that is \(\lim\limits_{x \to a - 0}f\left( x \right) \) \(= \lim\limits_{x \to a + 0}f\left( x \right) \). In this case,

\[\lim\limits_{x \to a}f\left( x \right) = \lim\limits_{x \to a - 0}f\left( x \right) = \lim\limits_{x \to a + 0}f\left( x \right).\]

Solved Problems

Example 1.

Using the \(\varepsilon-\delta-\) definition of limit, show that \[\lim\limits_{x \to 3} \left( {3x - 2} \right) = 7.\]

Solution.

Let \(\varepsilon \gt 0\) be an arbitrary positive number. Choose \(\delta = {\frac{\varepsilon }{3}}\). We see that if

\[0 \lt \left| {x - 3} \right| \lt \delta, \]

then

\[\left| {f\left( x \right) - L} \right| = \left| {\left( {3x - 2} \right) - 7} \right| = \left| {3x - 9} \right| = 3\left| {x - 3} \right| \lt 3\delta = 3 \cdot \frac{\varepsilon }{3} = \varepsilon .\]

Thus, by Cauchy definition, the limit is proved.

Example 2.

Using the \(\varepsilon-\delta-\) definition of limit, show that \[\lim\limits_{x \to 2} {x^2} = 4.\]

Solution.

For convenience, we will suppose that \(\delta = 1,\) that is

\[\left| {x - 2} \right| \lt 1.\]

Let \(\varepsilon \gt 0\) be an arbitrary number. Then we can write the following inequality:

\[\left| {{x^2} - 4} \right| \lt \varepsilon ,\;\; \Rightarrow \left| {x - 2} \right|\left| {x + 2} \right| \lt \varepsilon ,\;\; \Rightarrow \left| {x - 2} \right|\left( {x + 2} \right) \lt \varepsilon .\]

Since the maximum value of \(x\) is \(3\) (as we supposed above), we obtain

\[5\left| {x - 2} \right| \lt \varepsilon \;\;(\text{if } \left| {x - 2} \right| \lt 1),\;\; \text{or}\;\left| {x - 2} \right| \lt \frac{\varepsilon }{2}.\]

Then for any \(\varepsilon \gt 0\) we can choose the number \(\delta\) such that

\[\delta = \min \left( {\frac{\varepsilon }{2},1} \right).\]

As a result, the inequalities in the definition of limit will be satisfied. Therefore, the given limit is proved.

See more problems on Page 2.