A.4 Chapter 4

Let $y>x\geq 0$, so that $y^{2}+1>x^{2}+1$. Hence, $(y^{2}+1)^{-1}<(x^{2}+1)^{-1}$ and therefore

and so $f$ is increasing.

Suppose $x<y$. Since $\exp$ is surjective, there exists $a,b$ such that $x=\exp(a)$ and $y=\exp(b)$. Since $\exp$ is increasing, we must have $a<b$. Now, $\log(x)=\log(\exp(a))=a$ and $\log(y)=\log(\exp(b))=b$. Thus, since $a<b$, we have $\log(x)<\log(y)$.

Let $a$, $b\in\mathbb{R}$ with $a<b$. By the density of the rational numbers from Lemma 1.49, we know that there exists some $a<x_{1}<b$ with $x_{1}\in\mathbb{Q}$. By the density of the irrational numbers Lemma 1.53, we know that there exists some $x_{1}<y<b$ with $y\in\mathbb{R}\setminus\mathbb{Q}$. Finally, again by the density of the rational numbers from Lemma 1.49, we know that there exists some $y<x_{2}<b$ with $x_{2}\in\mathbb{Q}$. Observe that:

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  $a<x_{1}<y<b$ with $\chi_{\mathbb{Q}}(x_{1})=1>0=\chi_{\mathbb{Q}}(y)$ since $x_{1}\in\mathbb{Q}$ and $y\in\mathbb{R}\setminus\mathbb{Q}$. Hence $\chi_{\mathbb{Q}}$ is not nondecreasing on $(a,b)$.

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  $a<y<x_{2}<b$ with $\chi_{\mathbb{Q}}(y)=0<1=\chi_{\mathbb{Q}}(x_{2})$ since $x_{2}\in\mathbb{Q}$ and $y\in\mathbb{R}\setminus\mathbb{Q}$. Hence $\chi_{\mathbb{Q}}$ is not nonincreasing on $(a,b)$.

Thus, $\chi_{\mathbb{Q}}$ is not monotone on the interval $(a,b)$.

(i) Let $\delta>0$. If $x\in(-\infty,-\delta]\cup[\delta,\infty)$, then $|x|\geq\delta$. Consequently, $|r(x)|=1/|x|\leq 1/\delta$, and so $r$ is bounded above by $1/\delta$ and below by $-1/\delta$ on the set $(-\infty,-\delta]\cup[\delta,\infty)$.

(ii) Let $\delta>0$. Given any $M>0$, let $x\in\mathbb{R}$ satisfy $0<x<\min\{\delta,1/M\}$. Then $r(x)=1/x>M$ and so $M$ is not an upper bound for $r$ on $(-\delta,\delta)\setminus\{0\}$. However, since $M>0$ was chosen arbitrarily, it follows that $r$ is unbounded on $(-\delta,\delta)\setminus\{0\}$.

For $x\in\mathbb{R}$, we have

Let $\varepsilon>0$ be given and choose $\delta:=\varepsilon/5>0$. If $x\in\mathbb{R}$ satisfies $|x-1|<\delta$, then

Hence, by the $\varepsilon$-$\delta$ definition, the function $\ell$ is continuous at $1$.

We consider two cases:

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  If $x<0$, then $|f(x)-f(0)|=|x-0|=|x|$.

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  If $0\leq x<1$, then $|f(x)-f(0)|=|x^{3}-0|=|x|^{3}<|x|$.

In particular, for $x<1$ we have $|f(x)-f(0)|\leq|x|$.

Let $\varepsilon>0$ be given and choose $\delta:=\min\{\varepsilon,1\}>0$. Suppose $x\in\mathbb{R}$ satisfies $|x|<\delta$, so that $x<1$. By our earlier observation,

Thus, by the $\varepsilon$-$\delta$ definition, $f$ is continuous at $0$.

Let $a\in\mathbb{R}$. For $x\in\mathbb{R}$, we have

where the last step is by the triangle inequality. Now suppose $|x-a|<1$. Then another application of the triangle inequality yields

and so

Let $\varepsilon>0$ be given and choose

Suppose $x\in\mathbb{R}$ satisfies $|x-a|<\delta$. Then $|x-a|<1$ and so it follows by our earlier observations that

Thus, by the $\varepsilon$-$\delta$ definition, $p_{3}$ is continuous at $a$. Since $a\in\mathbb{R}$ was chosen arbitrarily, it follows that $p_{3}\colon\mathbb{R}\to\mathbb{R}$ is continuous.

Let $a\in(0,\infty)$. For $x\in(0,\infty)$ we have

Now suppose $|x-a|<|a|/2$. Then the triangle inequality yields

and so

Let $\varepsilon>0$ be given and choose

Suppose $x\in(0,\infty)$ satisfies $|x-a|<\delta$. Then $|x-a|<|a|/2$ and so it follows by our earlier observations that

Thus, by the $\varepsilon$-$\delta$ definition, $r$ is continuous at $a$.

Here is one suggested interpretation.

Fix $\varepsilon:=1/2$ and let $\delta>0$ be given. Let $x\in\mathbb{R}$ satisfy $0<x<\delta$, so that $|x|<\delta$ and

Hence, by definition, $f$ is discontinuous at $0$.

(i) We have already reduced to the case $\theta\in[0,\pi/2]$ and so $|\theta|=\theta$ and $|\tan\theta|=\tan\theta$.

The triangle $OAC$ in Figure 4.13(b) has area $(\tan\theta)/2$. On the other hand, the area of the sector $OAB$ in Figure 4.13(a) is $\theta/2$. Since the triangle contains the sector, $\theta/2\leq(\tan\theta)/2$ and the result follows.

(ii) By the double angle formula, $1-\cos\theta=2\sin^{2}(\theta/2)$. Thus, using the inequality $|\sin\theta|\leq|\theta|$, we have

as required.

Let $a\in\mathbb{Q}$ be a rational point, fix $\varepsilon:=1/2$ and let $\delta>0$ be given. By the density of the irrationals from Lemma 1.53, there exists some $x\in(a,a+\delta)$ with $x\in\mathbb{R}\setminus\mathbb{Q}$. In particular,

Thus, by definition, $\chi_{\mathbb{Q}}$ is discontinuous at $a$.

The function $g$ is continuous at $0$. Indeed, for $x\in\mathbb{R}\setminus\{0\}$, we have

since $|\sin(1/x)|\leq 1$. On the other hand, clearly $|g(x)-g(0)|\leq|x|$ also holds when $x=0$ (since both sides are $0$ in this case).

Let $\varepsilon>0$ be given and choose $\delta:=\varepsilon>0$. If $|x-0|<\delta$, then

Hence, by the $\varepsilon$-$\delta$ definition of continuity, $g$ is continuous at $0$.

We claim that $h$ continuous at $a=0$ and discontinuous at all other points.

To see $h$ is continuous at $a$, let $\varepsilon>0$ be given and choose $\delta:=\varepsilon$. If $x\in\mathbb{R}$ satisfies $|x|<\delta$, then

Hence, by the $\varepsilon$-$\delta$ definition, $h$ is continuous at $0$.

Now let $a\in\mathbb{R}\setminus\{0\}$ be a rational point; that is, $a\in\mathbb{Q}\setminus\{0\}$. We shall show $h$ is discontinuous at $a$. Indeed, let $\varepsilon:=|a|/2>0$ and $\delta>0$ be given. By the density of the irrationals from Lemma 1.53, we know that there exists some $x\in(a-\delta,a+\delta)$ with $x\in\mathbb{R}\setminus\mathbb{Q}$. Thus, $|x-a|<\delta$ and

Hence, by definition, $h$ is discontinuous at $a$.

Now let $a\in\mathbb{R}\setminus\{0\}$ be an irrational point. Using a similar argument to that above, we shall show $h$ is discontinuous at $a$. Indeed, again let $\varepsilon:=|a|/2>0$ and $\delta>0$ be given. Define $\delta_{0}:=\min\{\delta,|a|/2\}$. By the density of the rationals from Lemma 1.49, we know that there exists some $x\in(a-\delta_{0},a+\delta_{0})$ with $x\in\mathbb{Q}$. Thus, $|x-a|<\delta$ and, by the triangle inequality,

Hence, by definition, $h$ is discontinuous at $a$.

We know from Lemma 4.32 that $|\cos x-1|\leq x^{2}/2$ for all $x\in[-\pi/2,\pi/2]$. Consequently, if $0<|x|\leq\pi/2$, then

Let $\varepsilon>0$ be given and choose $\delta:=\min\{\pi/2,\varepsilon/2\}>0$. If $0<|x|<\delta$, then $0<|x|\leq\pi/2$ and it follows from our earlier observations that

Thus, by the $\varepsilon$-$\delta$ definition, $\frac{\cos x-1}{x}\to 0$ as $x\to 0$.

We claim that $\lim_{x\to 1}f(x)=0$. To see this, let $\varepsilon>0$ be given. We break the proof up into two parts.

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  Since $\sin\colon\mathbb{R}\to\mathbb{R}$ is continuous by Lemma 4.33 (and, in particular, continuous at the point $\pi$), there exists some $\eta_{1}>0$ such that

  $$|\sin(\pi x)-0|=|\sin(\pi x)-\sin(\pi)|<\varepsilon\qquad\text{if $x\in\mathbb{R}$ satisfies $|\pi x-\pi|<\eta_{1}$.}$$

  If we define $\delta_{1}:=\eta_{1}/\pi>0$, then it follows that

  $$|\sin(\pi x)-0|<\varepsilon\qquad\text{if $x\in\mathbb{R}$ satisfies $|x-1|<\delta_{1}$.}$$
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  For $x\in\mathbb{R}$, we have

  $$|x^{2}-1-0|=|x+1||x-1|$$

  If $|x-1|<1$, then it follows from the triangle inequality that

  $$|x+1|=|x-1+2|\leq|x-1|+2<3.$$

  Thus, in this case we have

  $$|x^{2}-1-0|=|x+1||x-1|<3|x-1|.$$

  Now choose $\delta_{2}:=\min\{1,\varepsilon/3\}>0$. If $x\in\mathbb{R}$ satisfies $|x-1|<\delta_{2}$, then $|x-1|<1$ and it follows from the above observations that

  $$|x^{2}-1-0|<3|x-1|<3\delta_{2}\leq\varepsilon.$$

Now set $\delta:=\min\{\delta_{1},\delta_{2}\}>0$. If $x\in\mathbb{R}$ satisfies $0<|x-1|<\delta$, then it follows from the above that $|f(x)-0|<\varepsilon$. Hence, by the $\varepsilon$-$\delta$ definition of a limit, $f(x)\to 0$ as $x\to 1$.

Finally, note that

and so the function $f$ is not continuous at $1$ by the limit characterisation of continuity from Lemma 4.42.

(i) There are many examples. For instance, let $f(x):=x$ for all $x\in\mathbb{R}$. Then we know that $\lim_{x\to 0}f(x)=0$, but $f$ is unbounded.

(ii) Let $I\subseteq\mathbb{R}$ be an interval, $a\in I$ and $f\colon I\setminus\{a\}\to\mathbb{R}$ and suppose $\ell:=\lim_{x\to a}f(x)$ exists. By the definition of a limit with $\varepsilon:=1$, there exists some $\delta_{0}>0$ such that

Define $M:=|\ell|+1$ and suppose $x\in(a-\delta_{0},a+\delta_{0})\cap I\setminus\{a\}$. Then, by the triangle inequality,

as required.

Let $\displaystyle\ell:=\lim_{x\to a}f(x)$ and $\displaystyle m:=\lim_{x\to a}g(x)$.

1. Let $\varepsilon>0$ be given. Since $f(x)\to\ell$ and $g(x)\to m$ as $x\to a$, by the $\varepsilon$-$\delta$ definition of a limit, there exists some $\delta_{1}$, $\delta_{2}>0$ such that

and

Choose $\delta:=\min\{\delta_{1},\delta_{2}\}>0$. By the triangle inequality,

Thus, by the $\varepsilon$-$\delta$ definition of a limit, $f(x)+g(x)\to\ell+m$ as $x\to a$.

2. By Exercise 4.48, since $f(x)\to\ell$ as $x\to a$, the function $f$ is locally bounded around $a$. More precisely, there exists some $\delta_{0}>0$ and some $M>0$ such that $|f(x)|\leq M$ for all $x\in I$ satisfying $0<|x-a|<\delta_{0}$.

Let $\varepsilon>0$ be given. Since $f(x)\to a$ and $g(x)\to b$ as $x\to a$, by the $\varepsilon$-$\delta$ definition of a limit, there exists some $\delta_{1}$, $\delta_{2}>0$ such that

and

Choose $\delta:=\min\{\delta_{0},\delta_{1},\delta_{2}\}>0$. By the triangle inequality,

Thus, by the $\varepsilon$-$\delta$ definition of a limit, $f(x)\cdot g(x)\to\ell\cdot m$ as $x\to a$.

3. We first show that $\displaystyle\lim_{x\to a}1/g(x)=1/m$.

Let $\varepsilon>0$. Since $g(x)\to b$ as $n\to\infty$ and $b\neq 0$, by the $\varepsilon$-$\delta$ definition of a limit, there exists some $\delta>0$ such that

Thus, if $x\in I$ satisfies $0<|x-a|<\delta$, then it follows by the triangle inequality that

and so

Hence, by the $\varepsilon$-$\delta$ definition of a limit, $1/g(x)\to 1/m$ as $x\to a$.

Since $f(x)\to\ell$ as $x\to a$ and $1/g(x)\to 1/m$ as $x\to a$, we can use the result from part 2 to conclude that $f(x)/g(x)\to\ell/m$ as $x\to a$, as required.

Let $I\subseteq\mathbb{R}$ be an interval and $f$, $g\colon I\to\mathbb{R}$ be continuous. Given $a\in I$, by the characterisation of continuity from Lemma 4.42, we know that

1) By Theorem 4.49 1) and (A.12), we know that

Hence, by Lemma 4.42, it follows that $f+g\colon I\to\mathbb{R}$ is continuous at $a$.

2) By Theorem 4.49 2) and (A.12), we know that

Hence, by Lemma 4.42, it follows that $f\cdot g\colon I\to\mathbb{R}$ is continuous at $a$.

3) Suppose $g(x)\neq 0$ for all $x\in I$. By Theorem 4.49 3) and (A.12), we know that

Hence, by Lemma 4.42, it follows that $(f/g)\colon I\to\mathbb{R}$ is continuous at $a$.

The above argument shows that $f+g$, $f\cdot g$ and, under the above additional hypothesis, $f/g$ are all continuous at $a\in I$. Since $a\in I$ was chosen arbitrarily, it follows that these functions are continuous.

We know from Lemma 4.33 that the functions $\sin\colon\mathbb{R}\to\mathbb{R}$ and $\cos\colon\mathbb{R}\to\mathbb{R}$ are both continuous. Furthermore, from Example 4.53, we know any polynomial function is continuous. It therefore follows from the limit laws that the functions

are both continuous. Finally, since $\sin^{4}x+5x^{2}+2\geq 2>0$ for all $x\in\mathbb{R}$, it again follows by the limit laws that

is continuous.

We know from Lemma 4.33 that $\sin\colon\mathbb{R}\to\mathbb{R}$ is continuous. We also know that $\sin x+2\geq 1>0$ for all $x\in\mathbb{R}$. Thus, from the limit laws, the function $x\mapsto(\sin x+2)^{-1}$ is continuous. Finally, again by Lemma 4.33, we know that $\cos\colon\mathbb{R}\to\mathbb{R}$ is continuous. Hence, the function $f\colon\mathbb{R}\to\mathbb{R}$ given by $f(x):=\cos\Big{(}\frac{1}{\sin x+2}\Big{)}$ for all $x\in\mathbb{R}$ is a composition of the continuous functions $\cos$ and $x\mapsto(\sin x+2)^{-1}$, and hence by the composition law (Theorem 4.55), $f$ is continuous.

(i) Note that $-1/n^{2}\to 0$ as $n\to\infty$. We know from (the borrowed!) Lemma 4.35 that the exponential function $\exp\colon\mathbb{R}\to\mathbb{R}$ is continuous. Thus, by the sequential continuity theorem,

(ii) We may write

We know that $n2^{-n}\to 0$ and $2^{2-n}\to 0$ as $n\to\infty$. Thus,

We know from (the borrowed!) Lemma 4.35 that the natural logarithm $\log\colon(0,\infty)\to\mathbb{R}$ is continuous. Thus, by the sequential continuity theorem,

For $x\in\mathbb{R}$, observe that

Thus, if we assume $x>8/7$, then

Let $\varepsilon>0$ and choose $R:=\max\{4/\varepsilon,8/7\}$. If $x>R$, then our earlier observations show that

Hence, by the $\varepsilon$-$R$ definition of a limit, $\tfrac{3x^{2}+5x+2}{2x^{2}+x+4}\to\tfrac{3}{2}$ as $x\to\infty$.

(i) For $x>0$ we may write

Let $\varepsilon>0$ be given. We know from Lemma 4.67 that there exists:

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  Some $R_{1}\geq 1$ such that $\exp(x)\geq 20\varepsilon^{-1}x^{3}$ for all $x>R_{1}$;

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  Some $R_{2}\geq 1$ such that $\exp(x)\geq 8\varepsilon^{-1}x^{2}$ for all $x>R_{2}$.

Let $R:=\max\{R_{1},R_{2}\}\geq 1$. Then for all $x>R$ we have

In particular,

Hence, by the $\varepsilon$-$R$ definition of a limit, $\frac{\exp(x)+10x^{3}+4x^{2}}{\exp(x)}\to 1$ as $x\to\infty$.

(ii) Let $\varepsilon>0$ be given. By Lemma 4.67 we know there exists some $S\geq 1$ such that $\log x\leq(\varepsilon x)^{1/100}$ for all $x>S$. Consequently,

Hence, by the $\varepsilon$-$R$ definition of a limit, $\frac{(\log x)^{100}}{x}\to 0$ as $x\to\infty$.