# Limit of a function

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 $f(x) = \frac{1}{x^2}$, $f: \mathbb{R} \to \mathbb{R} / \{0\}$.

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