Calculus

Limits and Continuity of Functions

Definition of Limit of a Function

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Problems 1-2
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Problems 3-5

Cauchy and Heine Definitions of Limit

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

The number \(L\) is called the limit of function \(f\left( x \right)\) as \(x \to a\) if and only if, for every \(\varepsilon \gt 0\) there exists \(\delta \gt 0\) 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 \(\varepsilon-\delta-\) 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 \(\left\{ {{x_n}} \right\}\), which has a limit at \(a,\) the sequence \(f\left( {{x_n}} \right)\) has a limit \(L.\) 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, i.e. \(\lim\limits_{x \to a – 0}f\left( x \right) \) \(= \lim\limits_{x \to a + 0}f\left( x \right) \). 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

Click on problem description to see solution.

 Example 1

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

 Example 2

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

 Example 3

Using the \(\varepsilon-\delta-\) definition of limit, find the number \(\delta\) that corresponds to the \(\varepsilon\) given with the following limit:
\[{\lim\limits_{x \to 7} \sqrt {x + 2} = 3,\;\;\;}\kern-0.3pt{\varepsilon = 0.2}\]

 Example 4

Prove that \(\lim\limits_{x \to \infty } {\large\frac{{x + 1}}{x}\normalsize} = 1\).

 Example 5

Prove that \(\lim\limits_{x \to \infty } {\large\frac{{2x – 3}}{{x + 1}}\normalsize} = 2\).

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 \(\delta = {\large\frac{\varepsilon }{3}\normalsize}\). 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 \(\delta = 1\), i.e.
\[\left| {x – 2} \right| \lt 1.\] 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
\[{5\left| {x – 2} \right| \lt \varepsilon \;\;(\text{if } \left| {x – 2} \right| \lt 1),\;\;}\kern-0.3pt
{\text{or}\;\left| {x – 2} \right| \lt \frac{\varepsilon }{2}.}
\] Then for any \(\varepsilon > 0\) 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.

Page 1
Problems 1-2
Page 2
Problems 3-5