Firstly, we will observe the limit of a function on a intuitive level. Consider the function

$$f(x) = \frac{9-x^2}{3+x}.$$

The domain of the function above is $\mathbb{R} / \{-3\}$, however, the value of the function can be calculated at any number which is ˝close˝ to $-3$.

From two tables above we can see that when the variable $x$ approaches $-3$, the value of the function $f$ approaches $6$.

We can simplify the given function and draw it in the coordinate plane:

$$f(x) = \frac{9-x^2}{3+x} = \frac{(3-x)(3+x)}{3+x} = 3-x$$

When the variable $x$ tends to $-3$, then we say that $6$ is the **limit** of the given function $f$.

Let $(x_n)$ be the sequence of the numbers which are located in the domain of the given function $f$. With this sequence is connected the sequence of the numbers $(f(x_n))$. That is, if we assume that the sequence $(x_n)$ approaches the number $a$, then we observe what is happening with the sequence $(f(x_n))$.

The following definition connects the limit of a sequence and the limit of a function.

Let $ I \subseteq \mathbb{R}$ be an open interval and $a \in I$. We say that a function $f: I/\{a\} \to \mathbb{R}$ has **a limit** $l$ **at the point** $a$ if for every sequence $(x_n)$ in $I/\{a\}$ the following is valid:

$$(\lim_{n \to \infty}x_n = a) \Rightarrow (\lim_{n \to \infty}f(x_n) = l).$$

We write:

$$l = \lim_{x \to a} f(x).$$

We can see that the convergence of the number $x$ to $a$ is equivalent to convergence of all sequences $(x_n)$ to $a$.

**One sided limits**

If we consider only sequences for which $x_n < a$, $\forall n \in \mathbb{N}$, then we say that the variable $x$ approaches $a$ from the left and we write $x \to a^{-}$. This means that the difference $x-a$ is negative. We write

$$\lim_{x \to a^{-}}f(x) = l.$$

On the other side, if we consider sequences for which $x_n > a$, $\forall n \in \mathbb{N}$, then we say that variable $x$ approaches $a$ from the right and we write $ x \to a^{+}$. This means that the difference $x-a$ is positive. We write

$$\lim_{ x \to a^{+}}f(x) = l.$$

Therefore, we can conclude that the limit of the function $f$ at the point $a$ exists if:

1.) exists the limit from the left,

2.) exists the limit from the right,

3.) limits from the left and from the right are equal.

**Cauchy definition**

The Cauchy definition of the limit of a function is independent of sequences.

Let $I \subseteq \mathbb{R}$ be an open interval, $a \in I$ and $f: I / \{a\} \to \mathbb{R}$. The limit of a function $f$ at the point $a$ exists and $\lim_{x \to a} f(x) = l$ if and only if

$$ (\forall \epsilon > 0) ( \exists \delta > 0) ((0 < | x – a| < \delta) \Rightarrow ( |f(x) – l | < \epsilon)).$$

**Cauchy definition of one sided limits**

Let $ I \subseteq \mathbb{R}$ be an open interval and $f: I/\{a\} \to \mathbb{R}$. The limit of the function $f$ at the point $a^{-}$ exists and $\lim_{x \to a^{-}} f(x) = l$ if and only if

$$(\forall \epsilon > 0) (\exists \delta > 0) (\forall x \in I)((0 < a – x < \delta) \Longrightarrow (|f(x) – l| < \epsilon)).$$

Let $ I \subseteq \mathbb{R}$ be an open interval and $f: I/\{a\} \to \mathbb{R}$. The limit of the function $f$ at the point $a^{+}$ exists and $\lim_{x \to a^{+}} f(x) = l$ if and only if

$$(\forall \epsilon > 0) (\exists \delta > 0) (\forall x \in I)((0 < x – a < \delta) \Longrightarrow (|f(x) – l| < \epsilon)).$$

The limit of the constant function is equal to:

$$\lim_{x \to a} c = c.$$

The limit of the identity function is equal to:

$$\lim_{x \to a} x = a.$$

The limit of a power is equal to:

$$\lim_{x \to a} x^n = a^n.$$

The limit of the $n$th root is equal to:

$$\lim_{x \to a} \sqrt[n]{x} = \sqrt[n] {a}.$$

**Properties of the limit of a function**

Let $I \subseteq \mathbb{R}$ be an open intervall, $a \in I$ and $f$ and $g$ two functions which have the limit at the point $a$, that is:

$$\lim_{x \to a}f(x) = l_1, \quad \lim_{x \to a} = l_2.$$

Then the following properties are valid:

1.) $$\lim_{x \to a}[f(x) \pm g(x) ] = l_1 \pm l_2 = \lim_{x \to a}f(x) \pm \lim_{x \to a}g(x),$$

2.) $\forall \alpha \in \mathbb{R}$ the function $f$ has the limit at the point $a$ and

$$lim_{x \to a} \alpha f(x) = \alpha \lim_{x \to a}f(x).$$

3.) $$\lim_{x \to a}[f(x) \cdot g(x)] = l_1 \cdot l_2 = \lim_{x \to a} f(x)) \cdot \lim_{x \to a} g(x) , $$

4.) $$\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{l_1}{l_2} = \frac{ \lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}, l_2\neq 0.$$

**Example 1**. Evaluate

$$\lim_{x \to 3}( x^2 + 3x -3).$$

**Solution**:

We know $$\lim_{x \to a} x^n = a ^n$$, $\lim_{x \to a}c=c$, and by using the property 1.) we have

$$\lim_{x \to 3} x^2 + 3x -3$$

$$= \frac{\lim_{x \to 2} 5}{ \lim_{x \to 2} x^3} = \lim_{x \to 3}x^2 + \lim_{x \to 3} 3x – \lim_{x \to 3}3$$

$$= 3^2 + 3 \cdot 3 – 3$$

$$ = 9 +9 -3$$

$$= 15.$$

**Example 2**. Evaluate

$$ \lim_{x \to 2} \frac{5}{x^3}.$$

**Solution**:

Since $\lim_{x \to 2} x^3 = 8 \neq 0$ we can use the property 3.):

$$\lim_{x \to 2} \frac{5}{x^3} $$

$$= \frac{5}{2^3} = \frac{5}{8}.$$

**Example 3**. Evaluate

$$ \lim_{ x \to 0} \frac{\sqrt{x^2+1} -2}{x^2 +1}.$$

**Solution**:

$$\lim_{ x \to 0} \frac{\sqrt{x^2+1} -2}{x^2 +1} = \frac{\lim_{x \to 0}(\sqrt{x^2+1} -2) }{\lim_{ x \to 0}(x^2+1)}$$

$$ = \frac{\sqrt{1} -2}{1}$$

$$ = -1. $$

Now we will show that

$$ \lim_{\alpha \to 0} \frac{\sin \alpha}{\alpha} = 1.$$

The function $f(\alpha) = \frac{\sin \alpha}{\alpha}$ is an odd function. Namely,

$$f( -\alpha) = \frac{\sin (-\alpha)}{-\alpha} = \frac{- \sin \alpha}{-\alpha} = \frac{\sin \alpha}{\alpha},$$

that is $f(-\alpha) = f(\alpha).$ Therefore, we will consider values of the function $f$ only for positive values of the variable $\alpha$.

On the unit circle we draw an angle $\alpha$, where $ 0 < \alpha < \frac{\pi}{2}$, $\alpha \in \mathbb{R}$.

The area of a sector ($P_s$) enclosed by two radii $|OA| = |OC| = 1$ and arc $\widehat{AB}$ is bigger than the area of a triangle $OAC$, and smaller than the area of a triangle $OAB$, that is

$$ P_{OAC} < P_s < P_{OAB}.$$

The triangle $OAC$ is an isosceles triangle with arms which lengths are equal to $|OA| = |OC| = 1$ and the angle between them $\alpha$. The length of the arc $\widehat{AB}$ is then equal to $\alpha$. A triangle $OAB$ is a right triangle where $|OA| = 1$ and $|AB| = \tan \alpha$. Therefore

$$\frac{\sin \alpha}{2} < \frac{\alpha}{2} < \frac{\tan \alpha}{2}.$$

The first inequality gives us

$$\frac{\sin \alpha}{2} < \frac{\alpha}{2} \Rightarrow \sin \alpha < x \Rightarrow \frac{\sin \alpha}{\alpha} < 1.$$

From the second inequality we have:

$$ \frac{\alpha}{2} < \frac{\tan \alpha}{2} \Rightarrow \alpha < \tan \alpha \Rightarrow \alpha < \frac{\sin \alpha}{\cos \alpha} \Rightarrow \cos \alpha < \frac{\sin \alpha}{\alpha}.$$

Finally:

$$ \cos \alpha < \frac{\sin \alpha}{ \alpha} < 1.$$

When the variable $\alpha$ approaches $0$, $\cos \alpha$ approaches $1$. Therefore, $\frac{\sin \alpha}{\alpha}$ also approaches $1$ when the variable $\alpha$ approaches $0$:

$$\lim_{\alpha \to 0} \frac{\sin \alpha}{\alpha} = 1.$$

**Example 4**. Evaluate

$$ \lim_{x \to 0} \frac{1 – \cos 2x}{x^2}.$$

**Solution**:

Since $ \cos 2x = \cos^2 x – \sin^2 x$ and from $\sin^2 x + \cos^2 x = 1$ we have

$$1 – \cos 2x = \sin^2 x + \cos^2 x – (\cos^2 x – \sin^2 x) = 2 \sin^2 x.$$

By the substitution $1 – \cos 2x = 2 \sin^2 x$ we have:

$$ \lim_{x \to 0} \frac{1 – \cos 2x}{x^2} = \lim_{x \to o} \frac{2 \sin^2 x}{x^2} = 2 \lim_{x \to 0}\left(\frac{\sin x}{x} \right)^2.$$

As we showed, $\lim_{x \to o} \frac{\sin x}{x} = 1$, therefore

$$\lim_{x \to 0} \frac{1 – \cos 2x}{x^2} = 2 \cdot 1^2 = 2.$$

**Limits at infinity**

A function $f: \left \langle a, + \infty \right \rangle \to \mathbb{R}$ has the limit $l \in \mathbb{R}$ at point $ + \infty$ if for every sequence $(x_n)$ in $\left \langle a, + \infty \right \rangle$ the following is valid:

$$\lim_{n \to +\infty}x_n = + \infty \Longrightarrow \lim_{n \to +\infty}f(x_n) =l.$$

We write:

$$\lim_{x \to + \infty} f(x) = l.$$

Similarly,

$$\lim_{x \to – \infty} f(x) = l.$$

**Example 5**. Evaluate

$$\lim_{x \to \infty} \frac{5 – 3x – x^2 }{5x^2 -1}.$$

**Solution**:

The highest power of $x$ appearing in the denominator is $x^2$. Therefore, we need to divide both the numerator and denominator by $x^2$. It follows

$$\lim_{x \to \infty} \frac{5 – 3x – x^2 }{5x^2 -1} = \lim_{x \to \infty} \frac{\frac{5}{x^2} – \frac{3}{x} – 1}{5 – \frac{1}{x^2}} = – \frac{1}{5}.$$

**Infinite limits**

Let $ I \subseteq \mathbb{R}$ be an open interval, $ a \in I$ and $f: I/\{a\} \to \mathbb{R}$. The function $f$ has the limit $ l = + \infty$ at point $a$ if for every sequence in $I/\{a\}$ the following is valid:

$$\lim_{x \to a} f(x) = + \infty.$$

Similarly,

$$\lim_{x \to a} f(x) = – \infty.$$

The properties of limits specified for limits of functions at point, also apply to limits at infinity and infinite limits.

**Example 6**. Consider the function $f(x) = \frac{1}{x^2}$, $f: \mathbb{R} \to \mathbb{R} / \{0\}$.

$$\lim_{x \to 0^{-}} \frac{1}{x^2} = – \infty, \quad \quad \lim_{x \to 0^{+}} \frac{1}{x^2} = + \infty.$$