🔬 Tutorial problems epsilon \(\epsilon\)

Note

This problems are designed to help you practice the concepts covered in the lectures. Not all problems may be covered in the tutorial, those left out are for additional practice on your own.

\(\epsilon\).1

Formulation

Consider the function

\[\begin{split} f(x)=\left\{\begin{array}{l} \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \text { if } x=1 \end{array}\right. \end{split}\]

(a) Does \(\lim _{x \rightarrow 5} f(x)\) exist? If so, what is it? Establish the validity of your answer formally using an epsilon-delta argument. If it exists, does it equal \(f(5)\) ? Is this function continuous at the point \(x=5\) ?

(b) Does \(\lim _{x \rightarrow 1} f(x)\) exist? If so, what is it? Establish the validity of your answer formally using an epsilon-delta argument. If it exists, does it equal \(f(1)\) ? Is this function continuous at the point \(x=1\) ?

(c) Is this function continuous on \(\mathbb{R}\)? Justify your answer.

Hint

Use examples in the lecture notes to establish \(\epsilon\)-\(\delta\) argument.

Solution

Consider the function

\[\begin{split} f(x)=\left\{\begin{array}{l} \frac{1}{x-1} \text { if } x \neq 1 \\ 0 \text { if } x=1 \end{array}\right. \end{split}\]

Part (a)

First of all, note that by the properties of the limits we have

\[ \lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5}\left(\frac{1}{x-1}\right)=\frac{\lim _{x \rightarrow 5} 1}{\lim _{x \rightarrow 5}(x-1)}=\frac{1}{4}=f(5) . \]

Thus we conclude that (i) \(\lim _{x \rightarrow 5} f(x)=\frac{1}{4}\), and (ii), the function is continuous at the point \(x=5\). However, we also want to show that \(\lim _{x \rightarrow 5} f(x)=\frac{1}{4}\) formally, using an epsilon-delta proof.

Denote \(d(a, b)\) the Euclidean distance between \(a\) and \(b\) in \(\mathbb{R}\). We have

\[ d(5, x)=\sqrt{(x-5)^{2}}=|x-5|, \]

and

\[ d(f(5), f(x))=\sqrt{(f(x)-f(5))^{2}}=|f(x)-f(5)|= |f(x)-\frac{1}{4}|. \]

Suppose that we want \(d(f(5), f(x))<\varepsilon\), where \(\varepsilon>0\). Any problems will be caused by small values of \(\varepsilon>0\). (Try and explain why this is the case.) Thus we can focus on the cases where

\[ d(f(5), f(x))= |f(x)-\frac{1}{4} |= |\frac{1}{(x-1)}-\frac{1}{4} | . \]

Note that

\[ d(f(5), f(x))<\varepsilon \Longleftrightarrow |\frac{1}{(x-1)}-\frac{1}{4} |<\varepsilon \]

This requires that both \(\frac{1}{(x-1)}-\frac{1}{4}<\varepsilon\) and \(-\left(\frac{1}{(x-1)}-\frac{1}{4}\right)<\varepsilon\).

We have

\[ \frac{1}{(x-1)}-\frac{1}{4}<\varepsilon \Longleftrightarrow \frac{1}{(x-1)}<\frac{1}{4}+\varepsilon \Longleftrightarrow \frac{1}{(x-1)}<\frac{1+4 \varepsilon}{4} . \]

This requires that

\[ x-1>\frac{4}{1+4 \epsilon} \Longleftrightarrow x>\frac{4}{(1+4 \epsilon)}+1 \Longleftrightarrow x>1+\frac{4}{(1+4 \epsilon)} \]

Note that

\[ -\left(\frac{1}{(x-1)}-\frac{1}{4}\right)<\varepsilon \Longleftrightarrow \frac{1}{(x-1)}-\frac{1}{4}>-\varepsilon \text {. } \]

We have

\[ \frac{1}{(x-1)}-\frac{1}{4}>-\varepsilon \Longleftrightarrow \frac{1}{(x-1)}>\frac{1}{4}-\varepsilon \Longleftrightarrow \frac{1}{(x-1)}>\frac{1-4 \varepsilon}{4} \]

This requires that

\[ x-1<\frac{4}{1-4 \epsilon} \Longleftrightarrow x<\frac{4}{(1-4 \epsilon)}+1 \Longleftrightarrow x<1+\frac{4}{(1-4 \epsilon)} \]

Thus we know that

\[ d(f(5), f(x))<\varepsilon \Longleftrightarrow 1+\frac{4}{(1+4 \epsilon)}<x<1+\frac{4}{(1-4 \epsilon)} . \]

Suppose that we want \(d(5, x)<\delta\). This requires that

\[ d(5, x)<\delta \Longleftrightarrow|x-5|<\delta \Longleftrightarrow-\delta<x-5<\delta \Longleftrightarrow 5-\delta<x<5+\delta \text {. } \]

Thus we want to choose \(\delta\) in such a way that

\[ 0<\delta \leqslant \min \left(4-\frac{4}{(1+4 \epsilon)},-4+\frac{4}{(1-4 \epsilon)}\right) \]

This can always be done, so we have

\[ \lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5}\left(\frac{1}{x-1}\right)=\frac{\lim _{x \rightarrow 5} 1}{\lim _{x \rightarrow 5}(x-1)}=\frac{1}{4}=f(5) . \]

Part (b)

It can be shown that \(\lim _{x \rightarrow 1} f(x)\) does not exist. This means that \(f(x)\) is discontinuous at the point \(x=1\).

Suppose that \(\lim _{x \rightarrow 1} f(x)=l\) for some \(l \in \mathbb{R}\). We will consider three distinct cases. The first case will involve \(l=0\). The second case will involve \(l>0\). The third case will involve \(l<0\).

Case 1: \(l=0\).

Suppose that \(\lim _{x \rightarrow 1} f(x)=0\). In order for \(\lim _{x \rightarrow 1} f(x)=0\), we need there to exist a function of \(\varepsilon, \delta: \mathbb{R}_{++} \longrightarrow \mathbb{R}_{++}\), such that, for each \(\varepsilon>0\), we have \(d(f(x), 0)<\varepsilon\) whenever \(d(x, 1)<\delta(\varepsilon)\). If we can find so much as a single \(\varepsilon>0\) for which no such \(\delta(\varepsilon)\) exists, then the limit does not exist. Note that

\[\begin{split} \begin{array}{ll} d(f(x), 0) &=\sqrt{(f(x)-0)^{2}} \\ & =|f(x)-0| \\ & =|f(x)| \\ & =\left\{\begin{array}{cc} \frac{1}{x-1}>0 & \text { if } x>1, \\ 0 & \text { if } x=1, \\ \frac{1}{1-x}>0 & \text { if } x<1 . \end{array}\right. \end{array} \end{split}\]

Let \(\varepsilon>0\). If \(x>1\), we have

\[ d(f(x), 0)<\varepsilon \Longleftrightarrow \frac{1}{x-1}<\varepsilon \Longleftrightarrow x-1>\frac{1}{\varepsilon} \Longleftrightarrow|x-1|>\frac{1}{\varepsilon} \]

because \(x-1>0\) when \(x>1\). On the other hand, if \(x<1\), we have

\[ d(f(x), 0)<\varepsilon \Longleftrightarrow \frac{1}{1-x}<\varepsilon \Longleftrightarrow 1-x>\frac{1}{\varepsilon} \Longleftrightarrow|x-1|>\frac{1}{\varepsilon}, \]

because \(1-x>0\) when \(x<1\). Recall that \(d(x, 1)=|x-1|\). Thus, when \(x \neq 1\) we know that \(d(f(x), 0)<\varepsilon\) if and only if \(d(x, 1)>\frac{1}{\varepsilon}\). Thus we know that, for any \(\varepsilon>0\), there is no \(\delta(\varepsilon)>0\) such that \(d(f(x), 0)<\varepsilon\) for all \(d(x, 1)<\delta(\varepsilon)\). This means that \(\lim _{x \rightarrow 1} f(x) \neq 0\). (This alone means that the function \(f(x)=\frac{1}{x-1}\) is not continuous at the point \(x=1\), because \(f(1)=0\).)

Case 2: \(l>0\).

Suppose that \(\lim _{x \rightarrow 1} f(x)=l>0\). In order for \(\lim _{x \rightarrow 1} f(x)=l>0\), we need there to exist a function of \(\varepsilon, \delta: \mathbb{R}_{++} \longrightarrow \mathbb{R}_{++}\), such that, for each \(\varepsilon>0\), we have \(d(f(x), l)<\varepsilon\) whenever \(d(x, 1)<\delta(\varepsilon)\). If we can find so much as a single \(\varepsilon>0\) for which no such \(\delta(\varepsilon)\) exists, then the limit does not exist. Note that

\[\begin{split} \begin{array}{ll} d(f(x), l) & =\sqrt{(f(x)-l)^{2}} \\ & =|f(x)-l| \\ & =\left\{\begin{array}{cl} \frac{1}{x-1}-l>0 & \text { if } \frac{1}{x-1}>l \text { and } x \neq 1, \\ l & \text { if } x=1, \\ l-\frac{1}{x-1}>0 & \text { if } \frac{1}{x-1}<l \text { and } x \neq 1 . \end{array}\right. \end{array} \end{split}\]

What happens when \(x=1\) here is irrelevant. What happens when \(x\) gets arbitrarily close to one is very relevant.

Let \(\varepsilon>0\). Suppose that \(f(x)>l>0\). Note that if \(f(x)>0\), we must have \(x>1\). In such a situation, we have

\[\begin{split} \begin{array}{l} d(f(x), l)<\varepsilon \Longleftrightarrow \frac{1}{x-1}-l<\varepsilon \Longleftrightarrow \frac{1}{x-1}<l+\varepsilon \\ \Longleftrightarrow x-1>\frac{1}{l+\varepsilon} \Longleftrightarrow|x-1|>\frac{1}{l+\varepsilon}, \end{array} \end{split}\]

because (i) \(x-1>0\) when \(x>1\), and (ii) \(l+\varepsilon>0\) when both \(\varepsilon>0\) and \(l>0\).

Let \(\varepsilon>0\). In fact, we will restrict attention to the subset of cases in which \(0<\varepsilon<l\). Suppose that \(0<f(x)<l\). Note that if \(f(x)>0\), we must have \(x>1\). In such a situation, we have

\[\begin{split} \begin{array}{l} d(f(x), l)<\varepsilon \Longleftrightarrow l-\frac{1}{x-1}<\varepsilon \Longleftrightarrow-\left(\frac{1}{x-1}\right)<-l+\varepsilon \Longleftrightarrow \frac{1}{x-1}>l-\varepsilon \\ \Longleftrightarrow 1>(x-1)(l-\varepsilon) \Longleftrightarrow x-1>\frac{1}{l-\varepsilon} \Longleftrightarrow|x-1|>\frac{1}{l-\varepsilon}>0, \end{array} \end{split}\]

because (i) \(x-1>0\) when \(x>1\), and (ii) \(l-\varepsilon>0\) when \(0<\varepsilon<l\).

Let \(\varepsilon>0\). In fact, we will restrict attention to the subset of cases in which \(0<\varepsilon<l\). Note that if \(x<1\), we must have \(f(x)<0<l\). This means that, in cases where \(x<1\), we have

\[\begin{split} \begin{array}{l} d(f(x), l)<\varepsilon \Longleftrightarrow l-\frac{1}{x-1}<\varepsilon \Longleftrightarrow-\left(\frac{1}{x-1}\right)<-l+\varepsilon \Longleftrightarrow \frac{1}{1-x}<l-\varepsilon \\ \Longleftrightarrow 1<(1-x)(l-\varepsilon) \Longleftrightarrow 1-x>\frac{1}{l-\varepsilon} \Longleftrightarrow|x-1|>\frac{1}{l-\varepsilon}>0, \end{array} \end{split}\]

because (i) \(x-1<0\) when \(x<1\), and (ii) \(l-\varepsilon>0\) when \(0<\varepsilon<l\).

We have thus established that when \(l>0\) and \(\varepsilon \in(0, l)\), we will have \(d(f(x), l)<\varepsilon\) if and only if \(d(x, 1)>\max \left(\frac{1}{l+\varepsilon}, \frac{1}{l-\varepsilon}\right)=\frac{1}{l-\varepsilon}>0\). Thus we know that, for any \(\varepsilon \in(0, l)\), there is no \(\delta(\varepsilon)>0\) such that \(d(f(x), l)<\varepsilon\) for all \(d(x, 1)<\delta(\varepsilon)\). This means that \(\lim _{x \rightarrow 1} f(x) \neq l\) for any \(l>0\).

Case 3: \(l<0\).

This case is left as an exercise. An appropriate variation on the arguments used for case 2 should work.

Combining the results from each of the three cases, we can conclude that the limit of the function as \(x\) approaches one does not exist. Since the limit of the function as \(x\) approaches one does not exist, we know that the function must be discontinuous at the point \(x=1\).

Part (c)

Since \(f(x)\) is discontinuous at the point \(x=1\), we know that it is a discontinuous function on \(\mathbb{R}\).

\(\epsilon\).2

Formulation

Consider the function

\[\begin{split} f(x)=\left\{\begin{array}{l} x^{2} \text { if } x \neq 0 \\ -50 \text { if } x=0 \end{array}\right. \end{split}\]

(a) Does \(\lim _{x \rightarrow 0} f(x)\) exist? If so, what is it? Try and establish the validity of your answer formally using an epsilon-delta argument. If it exists, does it equal \(f(0)\) ? Is this function continuous at the point \(x=0\) ?

(b) Is this function continuous?

Solution

Consider the function

\[\begin{split} f(x)=\left\{\begin{array}{l} x^{2} \text { if } x \neq 0 \\ -50 \text { if } x=0 \end{array}\right. \end{split}\]

Part (a)

It can be shown that \(\lim _{x \rightarrow 0} f(x)\) exists, but is equal to zero, so that

\[ \lim _{x \rightarrow 0} f(x) \neq f(0)=50 . \]

This means that \(f(x)\) is discontinuous at the point \(x=0\).

Denote \(d(a, b)\) the Euclidean distance between \(a\) and \(b\) in \(\mathbb{R}\). We have

\[ d(0, x)=\sqrt{(x-0)^{2}}=|x-0|=|x|, \]

and

\[ d(0, f(x))=\sqrt{(f(x)-0)^{2}}=|f(x)-0|=|f(x)| . \]

We want to find some \(\delta_{\varepsilon}>0\) for some arbitrary \(\varepsilon>0\), such that, whenever \(d(0, x)=|x|<\delta_{\varepsilon}\), we have \(d(0, f(x))=|f(x)|<\varepsilon\). We are interested in the behaviour of \(f(x)\) as \(x\) approaches zero here, not in the actual value taken by \(f(x)\) at the point \(x=0\). As such, we have

\[ d(0, f(x))=|f(x)|= |x^{2}|=x^{2} . \]

Consider some \(\varepsilon>0\). We want \(d(0, f(x))<\varepsilon\). Note that

\[ d(0, f(x))<\varepsilon \Longleftrightarrow x^{2}<\varepsilon \Longleftrightarrow \pm x<\sqrt{\varepsilon} \Longleftrightarrow-\sqrt{\varepsilon}<x<\sqrt{\varepsilon} . \]

Thus if we set \(\delta_{\varepsilon}=\sqrt{\varepsilon}\), then we know that \(d(0, f(x))<\varepsilon\) whenever \(d(0, x)<\delta_{\varepsilon}\). This means that \(\lim _{x \rightarrow 0} f(x)=0\).

Note that \(\lim _{x \rightarrow 0} f(x)=0 \neq-50=f(0)\). This means that the function \(f(x)\) is discontinuous at the point \(x=0\).

Part (b)

Since the point \(x=0\) belongs to the domain of the function \(f(x)\) and \(f(x)\) is discontinuous at the point \(x=0\), we know that it is a discontinuous function.

\(\epsilon\).3

Formulation

Consider the function

\[\begin{split} f(x)=\left\{\begin{array}{l} x^{2}+1 \text { if } x<2 \\ x+3 \text { if } x \geqslant 2 \end{array}\right. \end{split}\]

(a) Is this function continuous at the point \(x=2\) ? Justify your answer.

(b) Is this function continuous? Justify your answer.

Solution

Consider the function

\[\begin{split} f(x)=\left\{\begin{array}{l} x^{2}+1 \text { if } x<2 \\ x+3 \text { if } x \geqslant 2 \end{array}\right. \end{split}\]

Part (a)

We will show that

\[ \lim _{x \rightarrow 2} f(x)=5=f(2), \]

so that the function \(f(x)\) is continuous at the point \(x=2\).

Denote \(d(a, b)\) the Euclidean distance between \(a\) and \(b\) in \(\mathbb{R}\). We have

\[ d(2, x)=\sqrt{(x-2)^{2}}=|x-2|, \]

and

\[\begin{split} \begin{array}{ll} d(5, f(x)) & =\sqrt{(f(x)-5)^{2}}\\ &=|f(x)-5| \\ &= \left\{\begin{array}{l} |x^{2}+1-5| \text { if } x<2 ; \\ |x+3-5| \text { if } x \geqslant 2 ; \end{array}\right. \\ &= \left\{\begin{array}{l} |x^{2}-4| \text { if } x<2 ; \\ |x-2| \text { if } x \geqslant 2 . \end{array}\right. \end{array} \end{split}\]

Let \(\varepsilon>0\) and \(x<2\). Note that

\[ d(0, f(x))<\varepsilon \Longleftrightarrow |x^{2}-4 |<\varepsilon . \]

This requires that either (i) \(x^{2}-4<\varepsilon\), so that \(x^{2}<4+\varepsilon\), or (ii) \(-\left(x^{2}-4\right) < \varepsilon\), so that \(x^{2}>4-\varepsilon\). Thus we have

\[ d(0, f(x))<\varepsilon \Longleftrightarrow 4-\varepsilon<x^{2}<4+\varepsilon . \]

This means that, if \(\varepsilon>0\) and \(x<2\), then \(d(0, f(x))<\varepsilon\) requires that either

\[ \sqrt{4-\varepsilon}<x<\sqrt{4+\varepsilon} \]

or

\[ -\sqrt{4+\varepsilon}<x<-\sqrt{4-\varepsilon} . \]

Since we require that \(x>2\) in this case, we actually have

\[ d(0, f(x))<\varepsilon \Longleftrightarrow \]

(i) either \(\sqrt{4-\varepsilon}<x<2\) or (ii) \(-\sqrt{4+\varepsilon}<x<-\sqrt{4-\varepsilon}\).

Let \(\varepsilon>0\) and \(x \geqslant 2\). Note that

\[ d(0, f(x))<\varepsilon \Longleftrightarrow|x-2|<\varepsilon \Longleftrightarrow x<2+\varepsilon . \]

Since we require that \(x \geqslant 2\) in this case, we actually have

\[ d(0, f(x))<\varepsilon \Longleftrightarrow 2 \leqslant x<2+\varepsilon . \]

Upon combining our results, we can conclude that \(d(0, f(x))<\varepsilon\) if \(\sqrt{4-\varepsilon}<\) \(x<2\). Let

\[ \gamma_{\varepsilon}=2-\sqrt{4-\varepsilon} \]

and let \(\delta_{\varepsilon}=\min \left(\varepsilon, \gamma_{\varepsilon}\right)\). Note that \(\gamma_{\varepsilon}>0\), so that \(\delta_{\varepsilon}>0\). This ensures that \(d(0, f(x))<\varepsilon\) whenever \(d(2, x)<\delta_{\varepsilon}\). As such, we know that

\[ \lim _{x \rightarrow 2} f(x)=5=f(2), \]

so that the function \(f(x)\) is continuous at the point \(x=2\).

Part (b)

We have shown that the function

\[\begin{split} f(x)=\left\{\begin{array}{l} x^{2}+1 \text { if } x<2, \\ x+3 \text { if } x \geqslant 2 \end{array}\right. \end{split}\]

is continuous at the point \(x=2\). In order to show that the function is continuous, we also need to show that the function \(f(x)=x^{2}+1\) is continuous for all \(x<2\), and that the function \(f(x)=x+3\) is continuous for all \(x>2\). This can indeed be done, so that the function is indeed continuous. However, these additional steps are left as an exercise.

\(\epsilon\).4

Formulation

Use either the product rule or the quotient rule to find the first derivative of each of the following functions.

(a) \(y=x^{3}\left(x^{2}+4\right)\);

(b) \(y=(x-3)\left(x^{2}-5 x+7\right)\);

(c) \(y=\frac{\left(x^{3}+6 x^{2}-2\right)}{x^{4}}\);

(d) \(y=\left(x^{4}+3 x\right) x^{-6}\);

(e) \(y=5+x e^{x}\);

(f) \(y=x^{2} e^{x}+x \ln (x)\);

(g) \(y=x^{2}(1.1)^{x}\);

(h) \(y=5 e^{x} \ln (x)\);

(i) \(y=\left(x^{3}+2\right)(1.2)^{x}\); and

(j) \(y=x^{-\left(\frac{5}{6}\right)} e^{x}\).

[Shannon, 1995], p. 401

Solution

(a)

\[\begin{split} \begin{array}{ll} \frac{\mathrm{dy}}{\mathrm{dx}}&=\frac{\mathrm{df}}{\mathrm{dx}} \mathrm{g}(\mathrm{x})+\frac{\mathrm{dg}}{\mathrm{dx}} \mathrm{f}(\mathrm{x}) \\ &=3 x^{2}\left(x^{2}+4\right)+(2 x+0) x^{3}\\ &=3 x^{4}+12 x^{2}+2 x^{4} \\ &=5 x^{4}+12 x^{2} \end{array} \end{split}\]

(b)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x}&=\frac{d f}{d x} g(x)+\frac{d g}{d x} f(x) \\ & =(1)\left(x^{2}-5 x+7\right)+(2 x-5(1)+0)(x-3) \\ & =x^{2}-5 x+7+2 x^{2}-6 x-5 x+15 \\ & =3 x^{2}-16 x+22 \end{array} \end{split}\]

(c)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x}&=\frac{\frac{d f}{d x} g(x)-\frac{d g}{d x} f(x)}{[g(x)]^{2}} \\ \frac{d y}{d x} & =\frac{d}{d x}\left\{\frac{\left(x^{3}+6 x^{2}-2\right)}{x^{4}}\right\} \\ & =\frac{d}{d x}\left\{\left(x^{3}+6 x^{2}-2\right) x^{-4}\right\} \\ & =\left(\frac{d\left(x^{3}+6 x^{2}-2\right)}{d x}\right) x^{-4}+\left(\frac{d x^{-4}}{d x}\right)\left(x^{3}+6 x^{2}-2\right) \\ & =\left(3 x^{2}+12 x\right) x^{-4}+\left(-4 x^{-5}\right)\left(x^{3}+6 x^{2}-2\right) \\ & =3 x^{-2}+12 x^{-3}-4 x^{-2}-24 x^{-3}+8 x^{-5} \\ & =-x^{-2}-12 x^{-3}+8 x^{-5} . \end{array} \end{split}\]

(d)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x}&=\frac{\frac{d f}{d x} g(x)-\frac{d g}{d x} f(x)}{[g(x)]^{2}} \\ &=\frac{\left(4 x^{3}+3(1)\right) x^{6}-6 x^{5}\left(x^{4}+3 x\right)}{\left(x^{6}\right)^{2}}\\ & =\frac{4 x^{9}+3 x^{6}-6 x^{9}-18 x^{6}}{x^{12}} \\ & =\frac{-2 x^{9}-15 x^{6}}{x^{12}} \\ & =-2 x^{-3}-15 x^{-6} \end{array} \end{split}\]

(e)

\[ \frac{d y}{d x}=0+\frac{d f}{d x} g(x)+\frac{d g}{d x} f(x) =1 \cdot e^{x}+e^{x} \cdot x =e^{x}(1+x) \]

(f) In this example the product rule is used to find the derivative of both terms on the RHS

\[\begin{split} \begin{array}{ll} \frac{d y}{d x} & =\left[(2 x) e^{x}+\left(e^{x}\right) x^{2}\right]+\left[(1) \ln x+x \cdot \frac{1}{x}\right] \\ & =2 x e^{x}+x^{2} e^{x}+\ln x+1 \end{array} \end{split}\]

(g)

\[\begin{split} \begin{array}{ll} \frac{dy}{dx}&=\frac{\mathrm{df}}{\mathrm{dx}} \mathrm{g}(\mathrm{x})+\frac{\mathrm{dg}}{\mathrm{dx}} \mathrm{f}(\mathrm{x})\\ & =2 x(1.1)^{x}+\ln (1.1)(1.1)^{x} x^{2} \\ & =(1.1)^{x}\left(2 x+0.0953 x^{2}\right) \end{array} \end{split}\]

(h)

\[\begin{split} \begin{array}{ll} \frac{dy}{dx}&=\frac{\mathrm{df}}{\mathrm{dx}} \mathrm{g}(\mathrm{x})+\frac{\mathrm{dg}}{\mathrm{dx}} \mathrm{f}(\mathrm{x})\\ &=5 \cdot \mathrm{e}^{\mathrm{x}} \ln \mathrm{x}+5 \frac{1}{\mathrm{x}} \mathrm{e}^{\mathrm{x}}\\ &=5 \cdot \mathrm{e}^{\mathrm{x}}\left(\ln \mathrm{x}+\frac{1}{\mathrm{x}}\right) \end{array} \end{split}\]

(i)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x}&=\frac{d f}{d x} g(x)+\frac{d g}{d x} f(x)\\ & =\left(3 x^{2}+0\right)(1.2)^{x}+(\ln 1.2)(1.2)^{x}\left(x^{3}+2\right) \\ & =(1.2)^{x}\left[3 x^{2}+0.1823\left(x^{3}+2\right)\right] \\ & =(1.2)^{x}\left[0.1823 x^{3}+3 x^{2}+0.3646\right] \end{array} \end{split}\]

(j)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x}&=\frac{d f}{d x} g(x)+\frac{d g}{d x}(x)\\ &=-\frac{5}{6} x^{-11 / 6} e^{x}+e^{x} x^{-5 / 6}\\ &=e^{x}\left(x^{-5 / 6}-\frac{5}{6} x^{-11 / 6}\right) \end{array} \end{split}\]

\(\epsilon\).5

Formulation

Use the chain rule to find the first derivative of each of the following functions.

(a) \(y=(3 x+4)^{3}\);

(b) \(y=\left(2 x^{2}+6\right)^{4}\);

(c) \(y=(3-2 x)^{-2}\);

(d) \(y=\left(4+\ln \left(x^{2}\right)\right)^{-1}\);

(e) \(y=5+e^{x^{2}}\);

(f) \(y=10 x+e^{\ln (x)}\);

(g) \(y=\ln \left(x+x^{\frac{3}{2}}\right)\);

(h) \(y=\frac{10}{\left(1-5 e^{0.2 x}\right)}\);

(i) \(y=\sqrt{\frac{8}{x^{-1}}}\); and

(j) \(y=100-5 e^{-0.3 x}\).

[Shannon, 1995], p. 401

Solution

The chain rule or function of a function rule is

\[ \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x} \]

(a) \(y=(3 x+4)^{3}=u^{3}\) where \(u=3 x+4\)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x} & =\left(3 u^{2}\right)(3(1)+0) \\ & =9 u^{2}=9(3 x+4)^{2} \end{array} \end{split}\]

(b) \(y=\left(2 x^{2}+6\right)^{4}=u^{4}\) where \(u=2 x^{2}+6\)

\[\begin{split} \frac{d y}{d x}=\left(4 u^{3}\right)(2(2 x)+0)\\ =16 x u^{3}=16 x\left(2 x^{2}+6\right)^{3}\end{split}\]

(c) \(y=(3-2 x)^{-2}=u^{-2}\) where \(u=3-2 x\)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x} & =\left(-2 u^{-3}\right)(0-2(1)) \\ & =4 u^{-3}=4(3-2 x)^{-3} \end{array} \end{split}\]

(d) \(y=\left(4+\operatorname{ln} x^{2}\right)^{-1}=u^{-1}\) where \(u=4+\operatorname{ln} x^{2}\)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x} & =\left(-u^{-2}\right)\left(0+2 x \cdot \frac{1}{x^{2}}\right) \\ & =-\frac{2}{x}\left(u^{-2}\right)=-\frac{2}{x}\left(4+\ln x^{2}\right)^{-2} \end{array} \end{split}\]

(The chain rule was used to find the derivative of \(\operatorname{ln} x^{2}\).)

(e) As the derivative of 5 is 0 we only need use the chain rule to find the derivative of \(\mathrm{e}^{\mathrm{x}^{2}}\)

\(y =e^{x^{2}}=e^{u(x)} \text { where } u=x^{2} \\\)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x} & =e^{u(x)}(2 x) \\ & =2 x e^{x^{2}} \end{array} \end{split}\]

(f) The derivative of \(10x\) is \(10(1)=10\) so the chain rule is only needed to find the derivative of \(\mathrm{e}^{\ln x}\)

\(y =e^{\ln x}=e^{u(x)} \text { where } u=\ln x \\\)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x} & =e^{u(x)}\left(\frac{1}{x}\right) \\ & =\frac{1}{x} e^{\ln x}=\frac{1}{x} \cdot x=1 \end{array} \end{split}\]

The derivative of the original function is \(10+1\) or \(11\).

(g) \(\quad y=\operatorname{ln}\left(x+x^{3 / 2}\right)=\operatorname{ln} u\) where \(\mathrm{u}=x+x^{3 / 2}\)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x} & =\frac{1}{u} \cdot\left((1)+ \frac{3}{2} x^{1 / 2}\right) \\ & =\frac{(1+ \frac{3}{2} \sqrt{x})}{x+x^{3 / 2}} \end{array} \end{split}\]

(h) \(\quad y=\frac{10}{\left(1-e^{2 x}\right)}=10 u^{-1}\) where \(u=1-5 e^{2 x}\)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x} & =10\left(-u^{-2}\right)\left(0-5(0.2) e^{0.2 x}\right) \\ & =\frac{10 e^{0.2 x}}{\left(1-5 e^{0.2 x}\right)^{2}} \end{array} \end{split}\]

(i) \(y=\sqrt{8 x^{-1}}=u^{1 / 2}\), where \(u=8 x^{-1}\)

\[\begin{split} \begin{array}{ll} \frac{d y}{d x} & = \frac{d}{d x}\left\{\sqrt{\frac{8}{x^{-1}}}\right\} \\ & =\frac{d}{d x}\{\sqrt{8 x}\} \\ & =\frac{d}{d x}\left\{(8 x)^{\frac{1}{2}}\right\} \\ & =\left(\frac{d}{d(8 x)}\left\{(8 x)^{\frac{1}{2}}\right\}\right)\left(\frac{d}{d x}\{(8 x)\}\right) \\ & =\left\{\left(\frac{1}{2}\right)(8 x)^{-\left(\frac{1}{2}\right)}\right\}\{8\} \\ & =\left(\frac{8}{2}\right)(8 x)^{-\left(\frac{1}{2}\right)} \\ & =4\left(\frac{1}{(8 x)^{\frac{1}{2}}}\right) \\ & =4\left(\frac{1}{\sqrt{8 x}}\right) \\ & =\frac{4}{\sqrt{8 x}} \\ & =\frac{\sqrt{16}}{\sqrt{8 x}} \\ & =\sqrt{\frac{16}{8 x}} \\ & =\sqrt{\frac{2}{x}}, \end{array} \end{split}\]

assuming that \(x \neq 0\).

(j) The derivative of 100 is 0 so we only use the chain rule to find the derivative of \(-5 e^{-0.3 x}\)

\(y=-5 e^{-0.3 x}=-5 e^{u(x)} \text { where } u=-0.3 x\)

\[\begin{split} \begin{array}{ll} & \frac{d y}{d x}=-5 e^{u(x)}(-0.3) \\ &=1.5 e^{-0.3 x} \end{array} \end{split}\]

\(\epsilon\).6

Formulation

Consider a function \(f: X \longrightarrow \mathbb{R}\) where \(X \subseteq \mathbb{R}\). Suppose that \(\lim _{h \rightarrow 0}\left(\frac{f(a+h)-f(a)}{h}\right)\) exists. Show that this implies that \(f(x)\) is continuous at the point \(a \in X\).

Solution

Consider a function \(f: X \longrightarrow \mathbb{R}\) where \(X \subseteq \mathbb{R}\). Suppose that

\[ \lim _{h \rightarrow 0}\left(\frac{f(a+h)-f(a)}{h}\right) \]

exists. We want to show that this implies that \(f(x)\) is continuous at the point \(a \in X\).

This proof of this proposition is drawn from Ayers and Mendelson (2013, Chapter 8, Solved Problem 2).

First, note that

\[\begin{split} \begin{array}{ll} \lim _{h \rightarrow 0}(f(a+h)-f(a))&=\lim _{h \rightarrow 0}\left\{\left(\frac{h}{h}\right)(f(a+h)-f(a))\right\} \\ &=\lim _{h \rightarrow 0}\left\{h\left(\frac{f(a+h)-f(a)}{h}\right)\right\} \\ &=\lim _{h \rightarrow 0}(h) \lim _{h \rightarrow 0}\left(\frac{f(a+h)-f(a)}{h}\right) \\ &=(0)\left(\lim _{h \rightarrow 0}\left(\frac{f(a+h)-f(a)}{h}\right)\right)=0 . \end{array} \end{split}\]

Thus we have

\[ \lim _{h \rightarrow 0}(f(a+h)-f(a))=0 . \]

Now note that

\[\begin{split} \begin{array}{l} \lim _{h \rightarrow 0}(f(a+h)-f(a))=\lim _{h \rightarrow 0} f(a+h)-\lim _{h \rightarrow 0} f(a) \\ =\left(\lim _{h \rightarrow 0} f(a+h)\right)-f(a) . \end{array} \end{split}\]

Upon combining these two results, we obtain

\[ \left(\lim _{h \rightarrow 0} f(a+h)\right)-f(a)=0 \Longleftrightarrow \lim _{h \rightarrow 0} f(a+h)=f(a) . \]

Finally, note that

\[ \lim _{x \rightarrow a} f(x)=\lim _{h \rightarrow 0} f(a+h) . \]

Thus we have

\[ \lim _{x \rightarrow a} f(x)=f(a) \]

This means that \(f(x)\) is continuous at the point \(x=a\).

\(\epsilon\).7

Formulation

Approximate the function

\[ f(x) = \exp\left(\frac{1}{x}\right) + \ln(x) \]

around \(x = 1\) using the Taylor series.

Provide:

• The linear approximation

• The quadratic approximation

• The cubic approximation

Use each approximation to estimate the value of \(f(0)\). Is this an accurate estimate?

Hint

In other words, use as many terms of Tayor series to build an approximation by a linear function, quadratic and cubic polynomials.

Solution

We use the Taylor expansion formula about \(x = 1\):

\[ f(x) \approx f(1) + f'(1)(x - 1) + \frac{f''(1)}{2}(x - 1)^2 + \frac{f^{(3)}(1)}{2\cdot 3}(x - 1)^3 \]

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#Step 1: Evaluate function and derivatives at \(x = 1\)

\[ f(x) = \exp\left(\frac{1}{x}\right) + \ln(x) \]

\[ f(1) = e + \ln(1) = e \]

• First derivative:

  \[ f'(x) = -\frac{1}{x^2} e^{1/x} + \frac{1}{x} \quad \Rightarrow \quad f'(1) = -e + 1 = 1 - e \]

• Second derivative:

  \[ f''(x) = \left( \frac{2}{x^3} + \frac{1}{x^4} \right) e^{1/x} - \frac{1}{x^2} \quad \Rightarrow \quad f''(1) = 3e - 1 \]

• Third derivative:

  \[ f^{(3)}(x) = \left( -\frac{6}{x^4} - \frac{6}{x^5} - \frac{1}{x^6} \right) e^{1/x} + \frac{2}{x^3} \quad \Rightarrow \quad f^{(3)}(1) = -13e + 2 \]

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Linear Approximation

Using up to the first derivative:

\[ f(x) \approx f(1) + f'(1)(x - 1) \]

Substitute values:

\[ f(x) \approx e + (1 - e)(x - 1) \]

Compute approximation:

\[ f(0) \approx e + (1 - e)(-1) = 2e-1 \]

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Quadratic Approximation

Include up to the second derivative:

\[ f(x) \approx e + (1 - e)(x - 1) + \frac{3e - 1}{2}(x - 1)^2 \]

Compute approximation using previous approximation:

\[ f(0) \approx 2e-1 + \frac{3e - 1}{2} = \frac{7e-3}{2} \]

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Cubic Approximation

Include up to the third derivative:

\[ f(x) \approx e + (1 - e)(x - 1) + \frac{3e - 1}{2}(x - 1)^2 + \frac{-13e + 2}{6}(x - 1)^3 \]

Compute approximation using previous approximation:

\[ f(0) \approx \frac{7e-3}{2} - \frac{-13e + 2}{6} = \frac{34e -11}{6} \]

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All approximations are rather inaccurate because the function is undefined at \(x = 0\).

But as \(\lim_{x \to 0^+} f(x) = \infty\), so the approximated values are increasing with the order of the approximation.