2.3 Limits Continuity Derivatives 1

Goals

We provide, in this section, an intuitive definition of the three basic concepts: limit, continuity (direct substitution property), and derivative. We used these three concepts when solving the tangent line problem in the previous section.

 

2.3.1 The Concept of Limit for a Function

 

Definition 1. The limit of \(f(x)\) as \(x\) approaches \(a\) is equal to \(L\), denoted by
\[\lim_{x\to a} f(x)= L,\]
if the values \(f(x)\) are arbitrarily close to \(L\) for all \(x\) sufficiently close to \(a\), but \(x\neq a\).

Remark.

(1) It is important to note that \(f\) is not necessarily defined at \(a\).

 

(2) What is really necessary is that \(f\) be a function defined on an open interval \(I\) containing the point \(a\), but may or may not be defined at \(a\).

 

(3) A good example of this situation is the limit to find the slope of the tangent line to the graph of a function. We computed
\[\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.\]
Here, the quotient is not defined at \(x=a\), as in this case, the denominator vanishes.

Example 1. Give the function \(f\) whose graph is given in Figure 2.4, we have the following:

1. \(\displaystyle \lim_{x\to -3} f(x) = 0\), because the points on the graph of \(f\) are approaching the \(x\)-intercept \((-3,0)\) as \(x \to -3\), that is, the \(f(x)\)-values approach \(0\).

 

2. \(\displaystyle \lim_{x\to -2} f(x) \) DNE, does not exist, here the points on the graph of \(f\) approach the point \((-3,1)\) as \(x\to -2\) from the left, that is for values of \(x<-3\), while they approach \((-3,2)\) as \(x\to -2\) from the right, that is, for \(x>-3\).

 

3. \(\displaystyle \lim_{x\to 2} f(x) \) DNE, in this case, the values \(f(x)\to \infty\) as \(x\to 2\) from the right, and \(f(x)\to -\infty\) as \(x\to 2\) from the left.

 

4. \(\displaystyle \lim_{x\to 4} f(x)\) DNE, in this case, the values \(f(x)\to 4\) as \(x\to 4\) from the right, and \(f(x)\to 1\) as \(x\to 4\) from the left.
5. \(\displaystyle \lim_{x\to 5} f(x) = 2\).

As the above example shows, it is often necessary to study the values \(f(x)\) as \(x\) approaches from the left or from the right. These type of limits are called “one-sided” limits.

Definition 2. Let \(f\) be a function defined on an open interval \(I\) containing the point \(a\).

 

The right-hand limit of \(f(x)\) as \(x\) approaches \(a\) is equal to \(L\), denoted by
\[\lim_{x\to a^+} f(x)= L,\]
if the values \(f(x)\) are arbitrarily close to \(L\) for all \(x\) sufficiently close to \(a\), \(x>a\).

 

The left-hand limit of \(f(x)\) as \(x\) approaches \(a\) is equal to \(L\), denoted by

\[\lim_{x\to a^-} f(x)= L,\]
if the values \(f(x)\) are arbitrarily close to \(L\) for all \(x\) sufficiently close to \(a\), \(x < a\).

 

Remark. \(\displaystyle \lim_{x\to a} f(x)= L \Longleftrightarrow \lim_{x\to a^-} f(x)= L \;\; \mathrm{and} \;\; \lim_{x\to a^+} f(x)= L\).

 

Example 1. (Continuation.) Give the function \(f\) whose graph is given in Figure 2.5, explain the validity of each statement.

1. \(\displaystyle \lim_{x\to -4^+} f(x) = -1\);
2. \(\displaystyle \lim_{x\to -2^-} f(x) =1\);
3. \(\displaystyle \lim_{x\to -2^+} f(x) =2\).
4. \(\displaystyle \lim_{x\to 2^-} f(x) =-\infty\), DNE;
5. \(\displaystyle \lim_{x\to 2^+} f(x) =\infty\), DNE;
6. \(\displaystyle \lim_{x\to 4^-} f(x) =1\);
7. \(\displaystyle \lim_{x\to 4^+} f(x) =3\);
8. \(\displaystyle \lim_{x\to 5^-} f(x) = 2\);
9. \(\displaystyle \lim_{x\to 5^+} f(x) = 2\).

 

2.3.2 The Concept of Continuity of a Function

 

Figure 2.6(a) shows the graph of a function that is continuous at a point \(a\). As it should be expected, the graph has no gap, or breaks, or grows infinitely, at the point \(a\).

By contrast, the graphs in Figure 2.6 (b), (c) and (d) show the possible discontinuities that a function can have at a point \(a\):

(b) Removable discontinuity. The function can be redefined at \(a\) to render a continuous function.

(c) Jump discontinuity. The graph approaches different values from the left and from the right.

(d) Infinite discontinuity. The graph has a vertical asymptote at \(a\).

Definition 3. The function \(f\) is continuous at the point\(\pmb{a} \in \mathbb{R}\) if

(a) \(a \in D_f\);

(b) \(\displaystyle \lim_{x\to a} f(x)\) exists;

(c) \(\displaystyle \lim_{x\to a} f(x) = f(a)\).

Remark. Condition (c) is precisely the Direct Substitution Property (DSP).

Definition 4.

A function \(f\) is said to be continuous from the left at \(a\), if \(\displaystyle \lim_{x\to a^-} f(x) = f(a)\).

 

A function \(f\) is said to be continuous from the right at \(a\), if \(\displaystyle \lim_{x\to a^+} f(x) = f(a)\).

Remark. \(f\) is continuous at \(a\) \(\Longleftrightarrow\) \(f\) is continuous from the left and from the right at \(a\).

 

Example 2. Give the function \(f\) whose graph is given in Figure 2.7, explain the validity of each statement.

(a) \(f\) is discontinuous at \(-2\);

(b) \(f\) is continuous from the right at \(-2\);

(c) \(f\) has an infinite discontinuity at \(2\);

(d) \(f\) is continuous from the right at \(2\);

(e) \(f\) is continuous from the left at \(4\);

(f) \(f\) is discontinuous at \(4\).

 

Definition 5. Let \(f\) be a function defined on an open interval \(I\). \(f\) is said to be continuous on \(I\), if \(f\) is continuous at every point \(a \in I\).

 

Let \(f\) be a function defined on a closed interval \([c,\infty)\). \(f\) is said to be continuous on \(I\), if \(f\) is continuous at every point \(a \in (c,\infty)\), and \(f\) is continuous from the right at \(c\).

 

Let \(f\) be a function defined on a closed interval \((-\infty,c]\). \(f\) is said to be continuous on \(I\), if \(f\) is continuous at every point \(a \in (\infty,c)\), and \(f\) is continuous from the left at \(c\).

 

Let \(f\) be a function defined on a closed interval \([c,d]\). \(f\) is said to be continuous on \(I\), if \(f\) is continuous at every point \(a \in (c,d)\), \(f\) is continuous from the right at \(c\), and \(f\) is continuous from the left at \(d\).

 

Remark. Continuity on an interval intuitively means that the graph of \(f\) has no breaks at any point in the interval \(I\).

 

Example 2. (Continuation.) Give the function \(f\) whose graph is given in Figure 2.8, justify each statement.

(g) \(f\) is continuous on \((-4,-2)\);

(h) \(f\) is continuous on \([-2,2)\);

(i) \(f\) is continuous on \([2,4]\);

(j) \(f\) is continuous on \((4,6)\);

 

In what follows, we are going to use the following. A proof will be provided in the next section.

Proposition 1.

1. Polynomial functions are continuous on \(\mathbb{R}\).
2. Rational functions are continuous on their domains.
3. \(n\)-th root functions, \(f(x)=\sqrt[n]{x}\), are continuous on their domains.

Thus, for any function \(f\) in Proposition 1, the Direct Substitution Property is valid,
\[ \lim_{x\to a} f(x)= f(a), \;\; \forall \, a \in D_f.\]

Example 4.
By contrast, the function \(\displaystyle f(x)=\frac{|x|}{x}\) has a jump discontinuity at \(x=0\) , since
\begin{align*}
f[x]&=\frac{|x|}{x}
=\left [
\renewcommand{\arraystretch}{2}
\begin{array}{ll}
\displaystyle \frac{-x}{x}& x<0\\ \displaystyle \frac{x}{x}&x>0
\end{array}
\right.
=\left [
\renewcommand{\arraystretch}{1.5}
\begin{array}{ll}
-1& x<0\\ 1&x>0
\end{array}
\right.
\Longrightarrow
\left [
\renewcommand{\arraystretch}{2}
\begin{array}{l}
\displaystyle \lim_{x\to 0^-} f(x) = -1\\
\displaystyle \lim_{x\to 0^+} f(x) = 1
\end{array}
\right. .
\end{align*}
The graph in Figure 2.9 shows such a discontinuity.

2.3.3 The Concept of Differentiability of a Function

Definition 6. Let \(f\) be a function defined on an open interval \(I\) containing the point \(a\). \(f\) is said to be differentiable at \(a\), if

1. \(a \in D_f\), that is, \(f(a)\) is defined, and
2. \(\displaystyle \lim_{x\to a}\frac{f(x)-f(a)}{x-a}\) exists.

In this case, the limit is denoted by
\[f'(a) =\lim_{x\to a}\frac{f(x)-f(a)}{x-a},\]
and is called the derivative of \(f\) at \(a\).

Remark. (1) Condition (b) can be replaced by (b\(‘\)) \(\displaystyle \lim_{h\to 0}\frac{f(a+h)-f(a)}{h}\) exists.

 

(2) The differentiability of \(f\) and \(a\) has the following intuitive consequences:

• the graph must turn smoothly at \((a, f(a))\), as in Figure 2.10(a), that is,
• the graph must not have a corner at \((a, f(a))\), as in Figure 2.10(b), here the graph does not have a well defined tangent line, and
• the graph must at least be continuous at \(a\), failure to be continuous at \(a\) would lead to a situation as the cases in Figure 2.10(c) and (d) show.

(3) There is one more case where a function \(f\) may fail to be differentiable. This case corresponds to the situation where the function may have a vertical tangent line, such as the semicircle in Figure 2.10(e). In this case, \(\displaystyle \lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\infty\) (or \(-\infty\)), that is, the limit DNE.

 

As we have seen in 2.2.3, \(f'(a)\) yields the slope of the tangent line to the graph of \(f\) at \(a\). Thus, an equation for this tangent line is
\[y -f(a) = f'(a)(x-a) \;\;\; \text{ or} \;\;\; y =f(a) + f'(a)(x-a).\]

Definition 7. Let \(f\) be a function defined on an open interval \(I\). \(f\) is said to be differentiable on \(I\), if \(f\) is differentiable at each point \(a \in I\).

Example 3.
(a) Suppose \(f\) is a constant function, that is, \(f(x)=c\) for all \(x \in \mathbb{R}\), for some constant \(c\). Then,
\[f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a} = \lim_{x\to a}\frac{c-c}{x-a}=0
\mathrm{\ for \ all \ } a \in \mathbb{R},\]
that is, the derivative of a constant function is zero everywhere.

 

(b) Suppose \(f\) is a linear function, that is, \(f(x)= c x + d\) for all \(x \in \mathbb{R}\), for some constant numbers \(a,\ b\). Then,
\[f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a} = \lim_{x\to a}\frac{cx+d-(ca+d)}{x-a}=c
\mathrm{\ for \ all \ } a \in \mathbb{R},\]
that is, the derivative of a linear function \(f(x)=cx+d\) is constant and equal to \(c\) everywhere.

 

(c) Let \( f(x) = x^n\), \(n\in \mathbb{N}\), then, we already computed in Example 2 in 2.2.3,
\[f'(a)=\lim_{x \to a}\frac{f(x)-f(a)}{x-a} = n a^{n-1} \; \text{for all}\; a \in \mathbb{R}.\]
Thus, in particular, \(f\) is differentiable on \(\mathbb{R}\).

 

(d) Let \( f(x) = x^{-n}\), \(n\in \mathbb{N}\), then, it continues to be true that \(f'(a) =-n a^{-n-1}\) for \(a \in \mathbb{R}\), \(a\neq 0\), since
\[f'(a)=\lim_{x \to a}\frac{\frac{1}{x^n}-\frac{1}{a^n}}{x-a}
\underset{?}{=}-\lim_{x \to a}\frac{x^n-a^n}{(x-a)}\frac{1}{x^na^n}
\underset{?}{=}- n a^{n-1}\frac{1}{a^na^n} \underset{?}{=} -n a^{-n-1}.\]
Thus, in particular, \(f\) is differentiable on \((-\infty,0)\cup(0,\infty)\).

 

Remark. The result in part (b), should come as no surprise, given that \(c\) is the slope of the line which is the graph of \(f(x)=cx +d\), and the tangent line of a line is the line itself. The three rules obtained in the above example, will be constantly used throughout these notes.

 

Example 4. Given \(f(x)=\sqrt[4]{x}\), (a) use the definition of derivative to show that \(\displaystyle f'(a)= \frac{1}{4}\frac{1}{\sqrt[4]{a^3}}\), \(a>0\), (b) find an equation for the line tangent to the graph at the point with \(x= 16\).
(a) Solution 1. Let \(x \neq a\), \(x>0\) , and \(a>0\). We are going to show that
\[f'(a) = \lim_{x\to a} \frac{\sqrt[4]{x} – \sqrt[4]{a}}{x-a} = \frac{1}{4}\frac{1}{\sqrt[4]{a^3}} .\]
It is convenient to rewrite \(x = \left(\sqrt[4]{x}\right)^4\) and \(a = \left(\sqrt[4]{a}\right)^4\), and factor the denominator (this can be done because \(x>0\) and \(a>0\)).

\begin{align*}
\lim_{x\to a} \frac{\sqrt[4]{x} – \sqrt[4]{a}}{x-a}
& = \lim_{x\to a} \frac{\sqrt[4]{x} – \sqrt[4]{a}}{\left(\sqrt[4]{x}\right)^4-\left(\sqrt[4]{a}\right)^4} =\\
&= \lim_{x\to a} \frac{\sqrt[4]{x} – \sqrt[4]{a}}{\left(\sqrt[4]{x} – \sqrt[4]{a}\right)\left(\left(\sqrt[4]{x}\right)^3+\left(\sqrt[4]{x}\right)^2\left(\sqrt[4]{a}\right)+\left(\sqrt[4]{x}\right)\left(\sqrt[4]{a}\right)^2+\left(\sqrt[4]{a}\right)^3\right)}=\\
&= \lim_{x\to a} \frac{1}{\sqrt[4]{x^3}+\sqrt[4]{x^2a} + \sqrt[4]{xa^2} + \sqrt[4]{a^3}}
\underset{\mathrm{ DSP}}{=} \frac{1}{\sqrt[4]{a^3}+\sqrt[4]{a^2a} + \sqrt[4]{aa^2} + \sqrt[4]{a^3}}=\\
& =\frac{1}{4}\frac{1}{\sqrt[4]{a^3}}
\end{align*}

 

Solution 2. In this alternative solution, rather than factoring the denominator, we will multiply by the same expression both the numerator and the denominator to rewrite the difference quotient:
\begin{align*}
\lim_{x\to a} \frac{\sqrt[4]{x} – \sqrt[4]{a}}{x-a}
&= \lim_{x\to a} \frac{\sqrt[4]{x} – \sqrt[4]{a}}{x-a}\cdot
\frac{\sqrt[4]{x^3}+\sqrt[4]{x^2a} + \sqrt[4]{xa^2} + \sqrt[4]{a^3}}
{\sqrt[4]{x^3}+\sqrt[4]{x^2a} + \sqrt[4]{xa^2} + \sqrt[4]{a^3}}=\\
&\underset{?}{=} \lim_{x\to a} \frac{x-a}{(x-a)(\sqrt[4]{x^3}+\sqrt[4]{x^2a} + \sqrt[4]{xa^2} + \sqrt[4]{a^3})}
= \\
&= \lim_{x\to a} \frac{1}{\sqrt[4]{x^3}+\sqrt[4]{x^2a} + \sqrt[4]{xa^2} + \sqrt[4]{a^3}}
\underset{?}{=}\frac{1}{4}\frac{1}{\sqrt[4]{a^3}}.
\end{align*}

(b) The point on the graph is given by   \(\left (16, \sqrt[4]{16}\right ) = \left (16, 2\right )\).

The slope of the tangent line is given by   \(\displaystyle m =f’ (16) =\frac{1}{4}\frac{1}{\sqrt[4]{16^3}}= \frac{1}{32}\) (explain why).

An equation for the tangent line is given by   \(\displaystyle y = 2 + \frac{1}{32}(x – 16).\)

More generally, the following can be proven
\begin{equation}\label{ch02-eq09}
f(x)=\sqrt[n]{x}\Longrightarrow f'(x)=\frac{1}{n}\frac{1}{\displaystyle \sqrt[n]{x^{n-1}}}, \;\; x\in D_f, \;\; x\neq 0\tag{2.9}
\end{equation}

(see Exercise # 12).