2.4 Limits

Differentiation and integration, the two central concepts of calculus, hinge on the concept of limit. The purpose of this section is to provide the framework necessary to perform computations involving limits. We will stated some general results involving limits. However, the section will focus on computing limits related to polynomial and rational functions, and functions involving radicals. In Chapter 4, we shall study limits of trigonometric, inverse trigonometric, exponential and logarithmic functions.

2.4.1 Algebraic Operations Involving Limits

 
Theorem 1.
Let \(f\) and \(g\) be functions such that \(\displaystyle \lim_{x\to a}f(x)=L\) and \(\displaystyle \lim_{x\to a}g(x)=M\), then

1. \(\displaystyle \lim_{x\to a}(f(x)+g(x))= \lim_{x\to a}f(x)+ \lim_{x\to a}f(x))=L+M\);
2. \(\displaystyle \lim_{x\to a}(f(x)-g(x)) = \lim_{x\to a}f(x)-\lim_{x\to a}g(x)=L-M\);
3. \(\displaystyle \lim_{x\to a}f(x) \cdot g(x) = \lim_{x\to a}f(x)\cdot \lim_{x\to a}g(x)
   =L\cdot M\);
4. \(\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)} =
   \frac{\displaystyle \lim_{x\to a}f(x)}{\displaystyle \lim_{x\to a}g(x)}
   =\frac{L}{M}\), provided \(M\neq 0\).

Remarks. (1) We will use this theorem, even though we will not prove it.

(2) This theorem can be articulated in words. Under the right assumptions, we have

1. The limit of a sum is the sum of the limits.

2. The limit of a difference is the difference of the limits.

3. The limit of a product is the product of the limits.

4. The limit of a quotient is the quotient of the limits.

2.4.2 Evaluation of Limits

 
Here we restrict our attention to the evaluation of limits of polynomial, rational, and root functions.

Polynomials

 
Let \(f\) be a polynomial function. Then, because of the Direct Substitution Property (DSP) (see 2.2.2),
\[\lim_{x\to a}f(x) = f(a) \;\;\; \text{for all } \; a\in \mathbb{R}.\]

Rational Functions

 
Let \(\displaystyle f(x) = \frac{P(x)}{Q(x)}\) be a rational function. When computing the limit
\[\lim_{x\to a}\frac{P(x)}{Q(x)},\]
there are three cases to consider:

1. If \(Q(a)\neq 0\), then, by the DSP, \(\displaystyle \lim_{x\to a}\frac{P(x)}{Q(x)} = \frac{P(a)}{Q(a)}\).
2. If \(P(a)\neq 0\) and \(Q(a)=0\), then \(\displaystyle \left | \lim_{x\to a}\frac{P(x)}{Q(x)}\right | = \infty \).

   In this case, the graph of \(f\) has the line \(x = a\) as a vertical asymptote.

   Definition 1. A line \(x=a\) is called a vertical asymptote of the graph of a function
   \(f\) (or just of a function \(f\)) if at least one of the following four conditions holds.
   \[\lim_{x \to a^+} f(x) = \infty, \;\;\;\;\; \lim_{x \to a^+} f(x) = -\infty,\;\;\;\;\; \lim_{x \to a^-} f(x) = \infty, \;\;\;\;\;\lim_{x \to a^-} f(x) =- \infty\]
   (see Figure 2.13).

3. If \(P(a)= 0\) and \(Q(a)=0\), then the Direct Substitution Property yields the indeterminate form \(\displaystyle \frac{0}{0}\). Here, it is necessary to factor both \(P\) and \(Q\) and reduce \(\displaystyle f(x) = \frac{P(x)}{Q(x)}\) to lowest terms. The reduced rational function now falls into one of the two previous cases.

Example 1. Compute the limit. (a) \(\displaystyle \lim_{x\to 2}\frac{x^3-8}{x^4-16}\);     (b) \(\displaystyle \lim_{x\to 2}\frac{x^3-8}{x^2+4}\);     (c) \(\displaystyle \lim_{x\to -3}\frac{x – 3}{x^3+27}\).

Solution. (a) Note that the DSP yields the indeterminate form \(\displaystyle \frac{0}{0}\). We have
\[\lim_{x\to 2}\frac{x^3-8}{x^4-16}
=\lim_{x\to 2}\frac{(x-2)\left (x^2+2x+4\right )}{\left (x-2\right )\left (x^3+2x^2+4x+8\right )}
=\lim_{x\to 2}\frac{ x^2+2x+4 }{ x^3+2x^2+4x+8 }
\;\underset{\text{DSP}}{=}\;
\frac{12}{32}=\frac{3}{8}.
\]

(b) Since the denominator does not vanish at \(x=2\), we can immediately apply the DSP:
\[\lim_{x\to 2}\frac{x^3-8}{x^2 + 4} = \frac{0}{8}=0.\]

(c) Since the denominator vanishes at \(x=-3\), but the numerator does not, the graph of the function has a vertical asymptote.

 
Note that the numerator approaches \(-6\) as \(x\) approaches \(-3\), while the denominator approaches zero. Thus, the absolute value of the quotient grows to infinity. Furthermore, if \(x > -3\), then \(x^3+27 > 0 \), and, in this case, the quotient is negative. While if \(x < -3\), then \(x^3+27 < 0 \), and the quotient is positive. Thus, we have \[ \lim_{x\to -3^-}\frac{x-3}{x^3+27} = \infty\;\;\; \text{ and } \;\;\; \lim_{x\to -3^+}\frac{x-3}{x^3+27} = -\infty.\] Example 2. Compute the limit \(\displaystyle \lim_{x\to a}\frac{x^4-x^3+3x – a^4+a^3-3a}{x^2-a^2}\), \(a\neq 0\).

Solution. We use the properties of limits.

First, we decompose the rational expression whose limit is to be computed into a combination of three separate rational expressions. We now have that the original limit is the combination of the three separate limits (provided each limit exists).
\begin{align*}
\hspace{.3in}\hphantom{(*)}
\lim_{x\to a}\frac{x^4-x^3+3x – a^4+a^3-3a}{x^2-a^2}& =
\lim_{x\to a}\frac{x^4 – a^4 }{x^2-a^2} –
\lim_{x\to a}\frac{ x^3-a^3 }{x^2-a^2} +
\lim_{x\to a}\frac{ 3x – 3a}{x^2-a^2}\hspace{.3in}(*)
\end{align*}
Next, we verify that each separate limit exits and find its value.
 

(i) \(\displaystyle \lim_{x\to a}\frac{x^4 – a^4 }{x^2-a^2} = \lim_{x\to a}\frac{\left (x^2 – a^2\right )\left (x^2 + a^2\right ) }{x^2-a^2} = \lim_{x\to a} \left (x^2 + a^2 \right ) \;\underset{\text{DSP}}{=}\;2a^2\);

(ii) \(\displaystyle \lim_{x\to a}\frac{ x^3-a^3 }{x^2-a^2}
=\lim_{x\to a}\frac{ (x-a)\left (x^2+xa + a^2 \right )}{(x-a)(x+a)}
=\lim_{x\to a}\frac{ x^2+xa + a^2 }{ x+a } \;\underset{\text{DSP}}{=}\;\frac{3a}{2}\);

(iii) \(\displaystyle \lim_{x\to a}\frac{ 3x – 3a}{x^2-a^2} = \lim_{x\to a}\frac{ 3(x – a)}{(x-a)(x+a)} =\lim_{x\to a}\frac{ 3 }{ x+a } \;\underset{\text{DSP}}{=}\; \frac{3}{2a}, \;\; a\neq 0 \).

Since the three separate limits exist, from \((*)\) we have
\[\lim_{x\to a}\frac{x^4-x^3+3x – a^4+a^3-3a}{x^2-a^2} =
2a^2 – \frac{3a}{2} + \frac{3}{2a} = \frac{4a^3-3a^2+3}{2a}.\]

Roots

Example 3. Compute the limit. (a) \(\displaystyle \lim_{x\to 2}\frac{x^3-8}{\sqrt{x}-\sqrt{2}}\);    (b) \(\displaystyle \lim_{x\to -2}\frac{x^2+3x+2}{\sqrt[3]{x}+\sqrt[3]{2}}\).

Solution. (a) There are two ways to compute this limit (see Example 4 in 2.3.3).

Solution 1.
\begin{align*}
\lim_{x\to 2}\frac{x^3-8}{\sqrt{x}-\sqrt{2}} &=
\lim_{x\to 2}\frac{x^3-8}{\sqrt{x}-\sqrt{2}}\cdot\frac{\sqrt{x}+\sqrt{2}}{\sqrt{x}+\sqrt{2}} =
\lim_{x\to 2}\frac{(x-2)\left (x^2+2x+4\right )\left (\sqrt{x}+\sqrt{2}\right )}{x-2}
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
=\\ &=
\lim_{x\to 2} \left (x^2+2x+4\right )\left (\sqrt{x}+\sqrt{2}\right )
\underset{\text{DSP}}{=}(12)(2\sqrt{2})= 24\sqrt{2}.
\end{align*}
Solution 2.
\begin{align*}
\lim_{x\to 2}\frac{x^3-8}{\sqrt{x}-\sqrt{2}} &\underset{\hphantom{\text{DSP}}}{=}
\lim_{x\to 2}\frac{(x-2)\left (x^2+2x+4\right )}{\sqrt{x}-\sqrt{2}} =
\lim_{x\to 2}\frac{\left (\left (\sqrt{x}\right )^2 – \left (\sqrt{2}\right )^2\right )\left (x^2+2x+4\right )}{\sqrt{x} – \sqrt{2}}
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
=\\ &\underset{\hphantom{\text{DSP}}}{=}
\lim_{x\to 2}\frac{\left ( \sqrt{x} – \sqrt{2} \right )\left ( \sqrt{x} + \sqrt{2} \right )\left (x^2+2x+4\right )}{\sqrt{x} – \sqrt{2}}
\lim_{x\to 2}\left (\sqrt{x}+\sqrt{2}\right ) \left (x^2+2x+4\right )
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
=\\ &
\underset{\text{DSP}}{=}(2\sqrt{2})(12)= 24\sqrt{2}.
\end{align*}

(b) As in part(a), there are two ways to proceed.

Solution 1.
\begin{align*}
\lim_{x\to -2}\frac{x^2+3x+2}{\sqrt[3]{x}+\sqrt[3]{2}} &=
\lim_{x\to -2}\frac{(x+2)(x+1)}{\sqrt[3]{x}+\sqrt[3]{2}} =
\lim_{x\to -2}\frac{\left (\left (\sqrt[3]{x} \right )^3+ \left (\sqrt[3]{2} \right )^3\right )(x+1)}{\sqrt[3]{x}+\sqrt[3]{2}} =
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
=\\& =
\lim_{x\to -2}\frac{\left (\sqrt[3]{x} + \sqrt[3]{2} \right )\left (\left (\sqrt[3]{x} \right )^2- \sqrt[3]{x}\sqrt[3]{2} + \left (\sqrt[3]{2} \right )^2\right )(x+1)}{\sqrt[3]{x}+\sqrt[3]{2}}
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
=\\& =
\lim_{x\to -2} \left ( \sqrt[3]{x^2} – \sqrt[3]{2x} + \sqrt[3]{4} \right )(x+1)
\underset{\text{DSP}}{=}
\left ( \sqrt[3]{4} – \sqrt[3]{-4} + \sqrt[3]{4} \right )(-2+1)
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
=\\& =
-3\sqrt[3]{4}.
\end{align*}

Solution 2. Alternatively
\begin{align*}
\lim_{x\to -2}\frac{x^2+3x+2}{\sqrt[3]{x}+\sqrt[3]{2}} &=
\lim_{x\to -2}\frac{(x+2)(x+1)}{\sqrt[3]{x}+\sqrt[3]{2}} =
\lim_{x\to -2}\frac{(x+2)(x+1)}{\sqrt[3]{x}+\sqrt[3]{2}}
\cdot
\frac{ \left (\sqrt[3]{x} \right )^2- \sqrt[3]{x}\sqrt[3]{2} + \left (\sqrt[3]{2} \right )^2 }{ \left (\sqrt[3]{x} \right )^2- \sqrt[3]{x}\sqrt[3]{2} + \left (\sqrt[3]{2} \right )^2}
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
=\\& =
\lim_{x\to -2}\frac{(x+2)(x+1)\left (\left (\sqrt[3]{x} \right )^2- \sqrt[3]{x}\sqrt[3]{2} + \left (\sqrt[3]{2} \right )^2\right )}{x+2}
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
=\\& =
\lim_{x\to -2} (x+1)\left ( \sqrt[3]{x^2} – \sqrt[3]{2x} + \sqrt[3]{4} \right )
\underset{\text{DSP}}{=}
(-2+1) \left ( \sqrt[3]{4} – \sqrt[3]{-4} + \sqrt[3]{4} \right )
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
=\\& =
-3\sqrt[3]{4}.
\end{align*}

2.4.4 Side Limits

As seen in 2.4.2, side limits play a role when analyzing the behavior of a function near a vertical asymptote. Other contexts where side limits play a role deal with piecewise functions and functions involving absolute values.

Example 1. Find the value of \(a\) for which the limit \(\displaystyle \lim_{x\to x_0}f(x)\) exists (if any).

(a) \(f(x) = \left [\renewcommand{\arraystretch}{1.5}\begin{array}{ll}
\displaystyle \frac{x^3+1}{x+1}& \text{if }\; x < -1\\ ax+2 & \text{if } \; x > -1
\end{array}\right .\), \(x_0=-1\);      (b) \(g(x) = \left [\renewcommand{\arraystretch}{2}\begin{array}{ll}
\displaystyle \frac{\sqrt[3]{x^2} -1}{x^2-1}& \text{if }\; x < 1\\ \displaystyle \frac{x+a}{x+1} & \text{if } \; x > 1
\end{array}\right .\), \(x_0=1\).

Solution. (a)
\[\lim_{x\to -1^-}f(x)=\lim_{x\to -1^-}\frac{x^3+1}{x+1}= \lim_{x\to -1^-1}\frac{(x+1)\left (x^2-x+1\right )}{x+1}= \lim_{x\to -1^-1} \left (x^2-x+1\right ) \underset{\text{DSP}}{=} 3.\]
\[\lim_{x\to -1^+}f(x) = \lim_{x\to -1^+} ax+2 \underset{\text{DSP}}{=} -a+2.\]
Thus
\[\lim_{x\to -1^-}f(x)=\lim_{x\to -1^+} f(x) \;\; \Longleftrightarrow \;\; 3=-a+2 \;\; \Longleftrightarrow \;\; a=-1.\]
(b)
\[\lim_{x\to 1^-}\frac{\sqrt[3]{x^2} -1}{x^2-1} \underset{?}{=}
\lim_{x\to 1^-}\frac{\sqrt[3]{x^2} -1}{\left (\sqrt[3]{x^2} -1\right )\left (\sqrt[3]{x^4}+\sqrt[3]{x^2} +1\right )} = \lim_{x\to 1^-}\frac{1}{ \sqrt[3]{x^4}+\sqrt[3]{x^2} +1 } \underset{\text{DSP}}{=}\frac{1}{3}.
\]
\[ \lim_{x\to 1^+}\frac{x+a}{x+1} \underset{\text{DSP}}{=} \frac{1+a}{2}.\]
Thus
\[\lim_{x\to 1^-}g(x)=\lim_{x\to 1^+} g(x) \;\; \Longleftrightarrow \;\; \frac{1}{3}= \frac{1+a}{2} \;\; \Longleftrightarrow \;\; a=-\frac{1}{3}.\]

Example 2. Evaluate the limits (a) \(\displaystyle \lim_{x\to -1^-} \frac{1+x}{|1-x^2|} \);      (b) \(\displaystyle \lim_{x\to 2} \left (\frac{1}{|x-2|}-\frac{1}{x-2}\right ) \).

Solution. (a) Note that if \(x < -1\), then \(x+1 < 0\) and \(1 - x > 0\). Thus
\begin{align*}
\lim_{x\to -1^-} \frac{1+x}{|1-x^2|} & = \lim_{x\to -1^-} \frac{1+x}{|(1-x)(1+x)|}
= \lim_{x\to -1^-} \frac{1+x}{|1-x||1+x|} =
\vphantom{\frac{A}{\displaystyle \frac{A}{A}}}
\\
&= \lim_{x\to -1^-} \frac{1+x}{(1-x)(-(1+x))}
= \lim_{x\to -1^-} -\frac{1}{1-x} = -\frac{1}{2}.
\end{align*}
(b) We compute both side limits. If \(x < 2\), then \(x-2 < 0\), and \[\lim_{x\to 2^-} \left (\frac{1}{|x-2|}-\frac{1}{x-2}\right ) =\lim_{x\to 2^-} \left (\frac{1}{-(x-2)}-\frac{1}{x-2}\right ) =\lim_{x\to 2^-} -\frac{2}{x-2} = \infty;\] \[\lim_{x\to 2^+} \left (\frac{1}{|x-2|}-\frac{1}{x-2}\right ) \underset{?}{=} \lim_{x\to 2^+} \left (\frac{1}{x-2}-\frac{1}{x-2}\right )=0.\] It follows that the limit, \(\displaystyle \lim_{x\to 2} \left (\frac{1}{|x-2|}-\frac{1}{x-2}\right ) \vphantom{\frac{A}{\displaystyle \frac{a}{a}}}\), does not exist. However, the graph has a vertical asymptote at \(x=2\), as the Figure shows.

2.4.2 Algebraic Operations with Continuous Functions

An immediate consequence of Theorem 1 is the equivalent theorem for continuous functions.

Theorem 2.
Let \(f\) and \(g\) be continuous functions at a point \(a\), then
\(f+g\), \(f-g\), and \(f\cdot g\) are also continuous at \(a\). Furthermore,
\(f / g\) is also continuous at \(a\), provided \(g(a)\neq 0\).

Proof. We will prove that \(f / g\) is continuous at \(a\), whenever \(g(a)\neq 0\). The proofs for the other cases are similar.

According to Definition 3, we need to prove three things:

(a) \(a\in D_{f / g}\). This is true, since, by assumption, \(g(a)\neq 0\), thus \(f(a)/g(a)\) is well defined.

(b) \(\displaystyle \lim_{x\to a} \frac{f(x)}{g(x)}\) exists. This is immediate from Theorem 1(d) together with Definition 3(b).

(c) \(\displaystyle \lim_{x\to a} \frac{f(x)}{g(x)} =\frac{f(a)}{g(a)}\). This can be seen as follows:
\[
\lim_{x\to a} \frac{f(x)}{g(x)} \;
\underset{\mathrm{Th.1(d)}}{=} \;
\frac{\displaystyle \lim_{x\to a}f(x)}{\displaystyle \lim_{x\to a}g(x)}
\; \underset{\mathrm{Def.3(c)}}{=} \; \frac{f(a)}{g(a)}.
\]

We can now prove the following. Recall here that a rational function is a quotient of two polynomials.

Theorem 3.
1. Polynomials are continuous everywhere.

2. Rational functions are continuous everywhere in their domain.

Observation. In particular, this theorem states that if \(f\) is a polynomial or a rational function, then
\[\lim_{x\to a} f(x) = f(a) \;\; \mathrm{ for \; all } \;\; x \in D_f.\]

Proof
1. The proof is based on two basic results:

• If \(f\) is a constant function, \(f(x)=c\) for all \(x\), then \(f\) is continuous everywhere.This can be seen as follows:
  \[\lim_{x\to a} f(x)= \lim_{x\to a} c = c =f(a).\]
• If \(f(x)=x\) for all \(x\), then \(f\) is continuous everywhere.This can be seen as follows:
  \[\lim_{x\to a} f(x)= \lim_{x\to a} x = a= f(a).\]

Let’s see how these two facts are used to prove the theorem.

• \(f(x)=x^n\), \(n\in \mathbb{N}\) is continuous everywhere.This follows because, \(x^2= x \cdot x\) is a product of continuous functions, thus continuous. Then \(x^3 = x^2\cdot x\) is, again, a product of continuous functions, hence continuous, and so on.
• \(f_n(x) = a_n x^n\), \(n\in \mathbb{N}\), \(a_n\) a constant coefficient, is continuous everywhere.This follows because \(f\) is a product of continuous function, hence continuous.
• \(f(x) = a_nx^n + a_{n-1}x^{n-1}+\cdots + a_2x^2+ a_1x+a_0\) is continuous everywhere.This is true because it is a sum of \(n+1\) continuous functions.

2. Follows from 1 and the fact that the quotient of continuous functions is continuous wherever the denominator does not vanish, that is, in the domain of the rational function.

2.4.3 Intermediate Value Theorem

Here we state (without proof) an important theorem for continuous functions, and show how it is applied to prove the existence of solutions of polynomial/rational equations.

Theorem 4. (Intermediate Value Theorem Let \(f\) be a continuous function on the closed interval \([a,b]\), and assume that
\(N\) is a number between \(f(a)\) and \(f(b)\), then there exists a number \(c\) between \(a\) and \(b\) such that \(f(c) = N\).

Figure 2.13(a) IVT: assumptions

Figure 2.13(b) IVT: conclusion

Figure 2.13(c) Example 1

Figure 2.13: (a) IVT: assumptions (b) IVT: conclusion (c) Example 1

Observations. The Intermediate Value Theorem (IVT) is intuitively plausible. Figure 2.13(a) shows the assumptions of the theorem.

• One assumption is that \(N\) is a number between \(f(a)\) and \(f(b)\), thus, in particular, \(f(a)\neq f(b)\).
• Another assumption is that \(f\) is continuous on \([a,b]\).Thus, the graph of \(f\) is a curve with no breaks that connects the points
  \((a,f(a))\) and \((b,f(b))\), as shown in Figure 2.13(a).

As a consequence, we have that when drawing a horizontal line through the point \((0,N)\), because
\(N\) is between \(f(a)\) and \(f(b)\), this horizontal line must cross the graph of
\(f\) at some point \((c, f(c))\), as shown in Figure 2.13(b), \((c, f(c))\) must coincide with \((c,N)\). Thus, \(N=f(c)\). This is precisely the conclusion of the theorem.

An important application of the IVT is to prove the \textit{existence of solutions of equations}, such solutions are also called roots.

Example 1. Show that the equation \(x^4-2x^2-2x=2\) has at least one real root.

Solution. The question can be paraphrased as follows:
let \(f(x)=x^4-2x^2-2x\), then the problem is equivalent to show that there exists
\(c\in D_f\) such that \(f(c)=2\).

To apply the IVT, we need to we need to find an interval with endpoints \(a\) and \(b\) such that \(N=2\) lies between \(f(a)\) and \(f(b)\). This is normally done by trial and error (with some insight) by choosing the values of \(a\) and \(b\) in a way that it is easy to compute \(f(a)\) and \(f(b)\).

In our present case, we have \(f(0)=0<2\). Thus, we only need to find a number \(b\) such that \(f(b)>2\).

Try \(f(1)= 1-2-2=-3<2\), it does not work. Try \(f(2) = 2^4 – 2(2^2) – 2(2)= 16-12=4>2\), and we have \(f(0)=0<2<4=f(2)\). Thus we can choose \([a,b] = [0,2]\). Then, by the IVT, there exists a number \(c\in (0,2)\) such that \(f(c)=2\). Such a \(c\) is a root of the equation (see Figure 2.13(c)).
 
Remark. The fact that \(f(1)=-3<2<4=f(2)\), suggests that we can take
\([a,b] = [1,2]\), and hence that \(c\in (1,2)\), which gives a better estimate of \(c\).

2.4.4 Algebraic Operations Involving Derivatives

An immediate consequence of Theorem 1 is the equivalent theorem for differentiable functions. However, when computing the derivative of a product or a quotient, some additional considerations must be taken into account.

Theorem 5.
Let \(f\) and \(g\) be differentiable functions at a point \(a\), then

1. \(f+g\) is also differentiable at \(a\), and \((f+g)'(a)=f'(a)+g'(a)\).
2. \(f-g\) is also differentiable at \(a\), and \((f-g)'(a)=f'(a)-g'(a)\).
3. \(k f\) is also differentiable at \(a\), for any constant \(k\), and \((kf)'(a)=kf'(a)\).
4. Product Rule. \(f\cdot g\) is also differentiable at \(a\), and
   \[(f\cdot g)'(a)=f'(a)\cdot g(a)+f(a)\cdot g'(a).\]
5. \(\displaystyle \frac{1}{g}\) is also differentiable at \(a\), and
   \[\left (\frac{1}{g}\right )'(a)=-\frac{g'(a)}{(g(a))^2},\;\; \text{provided}\;\; g(a)\neq 0.\]
6. Quotient Rule. \(\displaystyle \frac{f}{g}\) is also differentiable at \(a\), and
   \[ \left (\frac{f}{g}\right )'(a)=\frac{g(a)f'(a)-f(a)g'(a)}{(g(a))^2},\;\; \text{provided}\;\; g(a)\neq 0.\]

Note. We will prove 1. Similar calculations prove 2 and 3. However, we will postpone the proof of the product rule, rule 5, and the quotient rule for later on. The goal at this point is for you to become fully familiar with these basic rules of differentiation and the proofs for 1–3.

Proof.

1.
\begin{align*}
\lim_{x\to a}\frac{(f+g)(x)-(f+g)(a)}{x-a}&
\underset{\hphantom{Th.\ref{ch02-th01}(a)}}{=} {\displaystyle \lim_{x\to a}}\frac{(f(x)+g(x))-(f(a)+g(a))}{x-a} =\\
&
\underset{\hphantom{Th.\ref{ch02-th01}(a)}}{=} {\displaystyle \lim_{x\to a}}\frac{(f(x)-f(a))+(g(x)-g(a))}{x-a} =\\
&\underset{Th.1(a)}{=}
\lim_{x\to a}\frac{f(x)-f(a)}{x-a} +
\lim_{x\to a}\frac{g(x)-g(a)}{x-a} = f'(a)+g'(a)
\end{align*}
Thus, \((f+g)'(a)=f'(a)+g'(a)\).

Remark.

• Rule 1 can be extended to finite sums of differentiable functions. For example, if \(f_1\), \(f_2\), and \(f_3\) are differentiable at \(a\), then \(f_1+f_2+f_3\) is also differentiable at \(a\).This can be seen as follows, first, \(f_1+f_2\) is differentiable at \(a\) because it is the sum of two differentiable functions at \(a\). Then, \(f_1+f_2+f_3\) is also differentiable at \(a\), because it is the sum of two differentiable functions at \(a\): \(f_1+f_2\) and \(f_3\). This argument can be extended for any finite sum of functions.
• Rule 4 can be extended to finite products of differentiable functions. For example, if \(f_1\), \(f_2\), and \(f_3\) are differentiable at \(a\), then \(f_1\cdot f_2 \cdot f_3\) is also differentiable at \(a\).The argument is similar to that in (1). Question: what would be the new product rule for three terms?

In addition, we have already proven the following rules of differentiation, where \(n\in\mathbb{Z}\), \(m\in \mathbb{N}\), \(m\geq 2\), \(a\) is a constant, and \(x\in D_{\sqrt[m]{x}}\), \(x\neq 0\).
\[\renewcommand{\arraystretch}{2}
\begin{array}{c|c||c|c||c|c}
f(x)&f'(x) &f(x)&f'(x) &f(x)&f'(x) \\ \hline
a& 0& x^n &nx^{n-1} & \sqrt[m]{x} &\displaystyle \frac{1}{m} \frac{1}{\sqrt[m]{x^{m-1}}} \\ \hline
\end{array}
\]

Theorem 5 together with the above rules lead to the following

Corollary 1.
1. Polynomial functions are differentiable everywhere in their domains.

2. Rational functions are differentiable everywhere in their domains.

3. \(f(x)=\sqrt[m]{x}\), \(m\in \mathbb{N}\), \(m\geq 2\), is differentiable everywhere in its domain.

Example 1. Compute the derivative of the following functions:

(a) \(f(x)=x^4-2x^2-2x-2\); (b) \(A(x) =(x^3+2x^2-1)\sqrt{x}\); (c) \(\displaystyle B(x) =\frac{x^3+2x-1}{\sqrt{x}}\).

Solution. (a) using 1, 2, and 3 repeatedly, the derivative of \(f\) is obtained by computing the derivative term by term as follows:
\(f'(x)=4x^3-4x-2\);

(b) applying the product rule:
\begin{align*}
A'(x) &=(x^3+2x^2 -1)’\sqrt{x}+(x^3+2x^2 -1)(\sqrt{x})’ =(3x^2+4x)\sqrt{x}+(x^3+2x^2 -1)\frac{1}{2\sqrt{x}} = \\
&=\frac{2(3x^2+4x)x+(x^3+2x^2 -1)}{2\sqrt{x}}=\frac{7 x^3+10 x^2-1}{2\sqrt{x}}
\end{align*}

(c) applying the quotient rule:
\begin{align*}
B'(x) =\frac{\sqrt{x}(3x^2+4x)-(x^3+2x^2 -1)\frac{1}{2\sqrt{x}}}{(\sqrt{x})^2}\underset{?}{=}\frac{5 x^3+6 x^2+1}{2x\sqrt{x}}.
\end{align*}

Example 2. Find the points where the graph of the function has a horizontal tangent line.

(a) \(\displaystyle f(x)= (x^2-6x+9)\left(x-2\right)-5x\); (b) \(\displaystyle g(x) = \frac{3x^2}{9-8x}\).

Solution. First of all, note the following:

horizontal tangent line at \((a,f(a)) \underset{?}{\Longleftrightarrow} f'(a)=0.\)

Thus, in each case, we need to compute \(f'(a)\), for all \(a\) where \(f\) is differentiable, and then solve \(f'(a)=0\) for such \(a\).

(a) Using the Product Rule
\[f'(a)= (2a-6)(a-2)+(a^2-3a+9)-5 \underset{?}{=}3 a^2-16 a+16 =(3 a-4)(a-4). \]
\[\Longrightarrow f'(a)=0 \Longleftrightarrow (3 a-4)(a-4) =0 \Longleftrightarrow a=\frac{4}{3} \;\text{or}\; a=4.\]
Thus, the graph has a horizontal tangent line at
\[ \left(\frac{4}{3},f\left(\frac{4}{3}\right)\right) =\left(\frac{4}{3},-\frac{230}{27}\right) \;\; \text{ and } \;\;
\left(4,f\left(4\right)\right) =\left(4,-18\right).\]

(b) Applying the Quotient Rule
\[g'(x)
=\frac{(9-8x)(6x)-(3x^2)(-8)}{(9-8x)^2}
=\frac{54x-48x^2+ 24x^2}{(9-8x)^2}
=\frac{6x (9-4 x)}{(9-8 x)^2} \]
\[\Longrightarrow g'(a)=0 \Longleftrightarrow \frac{6a (9-4 a)}{(9-8 a)^2} =0 \Longleftrightarrow 6a (9-4 a)=0 \Longleftrightarrow a=0 \;\text{or}\; a=\frac{9}{4}.\]
Thus, the points where the graph has a horizontal tangent line are
\[\left(0,g\left(0\right)\right) =\left(0,0\right)\;\; \text{and} \;\;
\left(\frac{9}{4},g\left(\frac{9}{4}\right)\right) =\left(\frac{9}{4},-\frac{27}{16}\right).\]

2.4.5 Mean Value Theorem for Derivatives

We end up this part by stating (without proof) an important theorem for differentiable functions,

Observations. The Mean Value Theorem (MVT) is intuitively plausible. Figure 2.14(a) shows the assumptions of the theorem.

• One assumption is that \(f\) is continuous on \([a,b]\). Thus, its graph has no gaps or breaks.
• Another assumption is that \(f\) is differentiable on \([a,b]\). Thus its graph has no corners, that is, it has a well defined tangent line at each point \((x,f(x))\) for \(a<x<b\).

Figure 2.14(a) MVT: assumptions

Figure 2.14(b) MVT: conclusion

Figure 2.14(c) MVT: plausibility

As a consequence, we have that when drawing the secant line connecting the points \(A(a,f(a))\) and \(B(b,f(b))\), and we slide it as in Figure 2.14(c), one of the sliding lines is bound to be tangential to the graph at a certain point \((c,f(c))\), for this point, the tangent line is parallel to the secant line \(AB\), Figure 2.14(b), thus, they have the same slope. This is precisely the conclusion of the theorem.

Example 3. Given the function \(\displaystyle f(x)=\frac{x^2}{x+2}\),

(a) show that \(f\) satisfies the hypothesis of the MVT on the interval \([1,2]\);

(b) find all numbers \(c\) that satisfy the conclusion of the MVT.

Solution. (a) Since \(f\) is a rational function, it is continuous on \(D_f = (-\infty, -2)\cup (-2,\infty)\). Thus, in particular, it is continuous on the interval \([1,2]\). Furthermore, \(f\) is differentiable in \(D_f\) (Corollary 1), thus, in particular, it is differentiable on \((1,2)\). Hence \(f\) satisfies the two assumptions of the MVT. It follows that there exists at least one number \(c\) in \((1,2)\) such that \(\displaystyle f'(c)=\frac{f(2)-f(1)}{2-1}\).

(b) We need to find the numbers \(c \in (1,2)\) such that \(\displaystyle f'(c)=\frac{f(2)-f(1)}{2-1}\). Since
\(\displaystyle f'(x)=\frac{x (x+4)}{(x+2)^2}\) (verify this),
\[f'(c)=\frac{f(2)-f(1)}{2-1} =\frac{1-\frac{1}{3}}{2-1}=\frac{2}{3} \Longleftrightarrow \frac{c (c+4)}{(c+2)^2} =\frac{2}{3} \Longleftrightarrow\]
\[\underset{?}{\Longleftrightarrow} c = 2 (-1 \pm \sqrt{3}) \Longrightarrow c=2 (-1 + \sqrt{3}) \in (1,2).\]
Note that \(2 (-1 – \sqrt{3}) \) is ruled out because it is not in the interval \((1,2)\).