# Calculus

## Limits and Continuity of Functions # Definition of Limit of a Function

### 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, 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

Click or tap a problem to see the 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 $$\varepsilon \gt 0$$ 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.

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Problems 1-2
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Problems 3-5