극한의 엄밀한 정의(ε-δ)

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-4|<\delta$, $|f(x)-15|<\epsilon$.

Consider $|f(x)-15|<\epsilon$, keeping in mind we want to make a statement about $|x-4|$:

$$\begin{gathered} |f(x) -15 | < \epsilon \\ |x^2+x-5 -15 |<\epsilon \\ | x^2+x-20 | < \epsilon \\ | x-4 |\cdot|x+5| < \epsilon \\ | x-4 | < \epsilon/|x+5| \\ \end{gathered}$$

Since $x$ is near 4, we can safely assume that, for instance, $3<x<5$. Thus

$$\begin{gathered} 3+5<x+5<5+5 \\ 8 < x+5 < 10 \\ \frac{1}{10} < \frac{1}{x+5} < \frac{1}{8} \\ \frac{\epsilon}{10} < \frac{\epsilon}{x+5} < \frac{\epsilon}{8} \\ \end{gathered}$$

Let $\delta =\frac{\epsilon}{10}$. Then:

$$\begin{gathered} |x-4|<\delta \\ |x-4| < \frac{\epsilon}{10}\\ |x-4| < \frac{\epsilon}{x+5}\\ |x-4|\cdot|x+5| < \frac{\epsilon}{x+5}\cdot|x+5|\\ \end{gathered}$$

Assuming $x$ is near 4, $x+5$ is positive and we can drop the absolute value signs on the right.

$$\begin{gathered} |x-4|\cdot|x+5| < \frac{\epsilon}{x+5}\cdot(x+5)\\ |x^2+x-20| < \epsilon\\ |(x^2+x-5) -15| < \epsilon, \end{gathered}$$

which is what we wanted to prove.

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-1|<\delta$, $|f(x)-6|<\epsilon$.

Consider $|f(x)-6|<\epsilon$, keeping in mind we want to make a statement about $|x-1|$:

$$\begin{gathered} |f(x) -6 | < \epsilon \\ |(2x^2+3x+1)-6 |<\epsilon \\ | 2x^2+3x-5 | < \epsilon \\ | 2x+5 |\cdot|x-1| < \epsilon \\ | x-1 | < \epsilon/|2x+5| \\ \end{gathered}$$

Since $x$ is near 1, we can safely assume that, for instance, $0<x<2$. Thus

$$\begin{gathered} 0+5<2x+5<4+5 \\ 5 < 2x+5 < 9 \\ \frac{1}{9} < \frac{1}{2x+5} < \frac{1}{5} \\ \frac{\epsilon}{9} < \frac{\epsilon}{2x+5} < \frac{\epsilon}{5} \\ \end{gathered}$$

Let $\delta =\frac{\epsilon}{9}$. Then:

$$\begin{gathered} |x-1|<\delta \\ |x-1| < \frac{\epsilon}{9}\\ |x-1| < \frac{\epsilon}{2x+5}\\ |x-1|\cdot|2x+5| < \frac{\epsilon}{2x+5}\cdot|2x+5|\\ \end{gathered}$$

Assuming $x$ is near 1, $2x+5$ is positive and we can drop the absolute value signs on the right.

$$\begin{gathered} |x-1|\cdot|2x+5| < \frac{\epsilon}{2x+5}\cdot(2x+5)\\ |2x^2+3x-5| < \epsilon\\ |(2x^2+3x+1) -6| < \epsilon, \end{gathered}$$

which is what we wanted to prove.

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-2|<\delta$, $|f(x)-7|<\epsilon$.

Consider $|f(x)-7|<\epsilon$, keeping in mind we want to make a statement about $|x-2|$:

$$\begin{gathered} |f(x) -7 | < \epsilon \\ |x^3-1 -7 |<\epsilon \\ | x^3-8 | < \epsilon \\ | x-2 |\cdot|x^2+2x+4| < \epsilon \\ | x-3 | < \epsilon/|x^2+2x+4| \\ \end{gathered}$$

Since $x$ is near 2, we can safely assume that, for instance, $1<x<3$. Thus

$$\begin{gathered} 1^2+2\cdot1+4<x^2+2x+4<3^2+2\cdot3+4 \\ 7 < x^2+2x+4 < 19 \\ \frac{1}{19} < \frac{1}{x^2+2x+4} < \frac{1}{7} \\ \frac{\epsilon}{19} < \frac{\epsilon}{x^2+2x+4} < \frac{\epsilon}{7} \\ \end{gathered}$$

Let $\delta =\frac{\epsilon}{19}$. Then:

$$\begin{gathered} |x-2|<\delta \\ |x-2| < \frac{\epsilon}{19}\\ |x-2| < \frac{\epsilon}{x^2+2x+4}\\ |x-2|\cdot|x^2+2x+4| < \frac{\epsilon}{x^2+2x+4}\cdot|x^2+2x+4|\\ \end{gathered}$$

Assuming $x$ is near 2, $x^2+2x+4$ is positive and we can drop the absolute value signs on the right.

$$\begin{gathered} |x-2|\cdot|x^2+2x+4| < \frac{\epsilon}{x^2+2x+4}\cdot(x^2+2x+4)\\ |x^3-8| < \epsilon\\ |(x^3-1) - 7| < \epsilon, \end{gathered}$$

which is what we wanted to prove.

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-2|<\delta$, $|f(x)-5|<\epsilon$. However, since $f(x)=5$, a constant function, the latter inequality is simply $|5-5|<\epsilon$, which is always true. Thus we can choose any $\delta$ we like; we arbitrarily choose $\delta =\epsilon$.

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-0|<\delta$, $|f(x)-0|<\epsilon$.

Consider $|f(x)-0|<\epsilon$, keeping in mind we want to make a statement about $|x-0|$ (i.e., $|x|$):

$$\begin{gathered} |f(x) -0 | < \epsilon \\ |e^{2x}-1 |<\epsilon \\ -\epsilon< e^{2x}-1 < \epsilon \\ 1-\epsilon< e^{2x} < 1+\epsilon \\ \ln (1-\epsilon) < 2x < \ln (1+\epsilon) \\ \frac{\ln (1-\epsilon)}{2} < x < \frac{\ln (1+\epsilon)}{2} \\ \end{gathered}$$

Let $\delta = \min\left\{\left|\frac{\ln(1-\epsilon)}{2}\right|,\frac{\ln(1+\epsilon)}{2}\right\}=\frac{\ln(1+\epsilon)}{2}.$

Thus:

$$\begin{gathered} |x| < \delta \\ |x| <\frac{\ln(1+\epsilon)}{2}<\left|\frac{\ln(1-\epsilon)}{2}\right| \\ \frac{\ln(1-\epsilon)}{2} < x < \frac{\ln(1+\epsilon)}{2}\\ \ln(1-\epsilon)< 2x < \ln(1+\epsilon)\\ 1-\epsilon < e^{2x} < 1+\epsilon\\ -\epsilon < e^{2x}-1 < \epsilon\\ |e^{2x}-1-(0)| < \epsilon, \end{gathered}$$

which is what we wanted to prove.

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-1|<\delta$, $|f(x)-1|<\epsilon$.

Consider $|f(x)-1|<\epsilon$, keeping in mind we want to make a statement about $|x-1|$:

$$\begin{gathered} |f(x) -1 | < \epsilon \\ |1/x-1 |<\epsilon \\ | (1-x)/x | < \epsilon \\ | x-1 |/|x| < \epsilon \\ | x-1 | < \epsilon\cdot|x| \\ \end{gathered}$$

Since $x$ is near 1, we can safely assume that, for instance, $1/2<x<3/2$. Thus $\epsilon/2 < \epsilon\cdot x$.

Let $\delta =\frac{\epsilon}{2}$. Then:

$$\begin{gathered} |x-1|<\delta \\ |x-1| < \frac{\epsilon}{2}\\ |x-1| < \epsilon\cdot x\\ |x-1|/x < \epsilon\\ \end{gathered}$$

Assuming $x$ is near 1, $x$ is positive and we can bring it into the absolute value signs on the left.

$$\begin{gathered} |(x-1)/x| < \epsilon\\ |1-1/x| < \epsilon\\ |(1/x) -1| < \epsilon, \end{gathered}$$

which is what we wanted to prove.

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-0|<\delta$, $|f(x)-0|<\epsilon$. In simpler terms, we want to show that when $|x|<\delta$, $|\sin x| < \epsilon$.

Set $\delta = \epsilon$. We start with assuming that $|x|<\delta$. Using the hint, we have that $|\sin x | < |x| < \delta = \epsilon$. Hence if $|x|<\delta$, we know immediately that $|\sin x| < \epsilon$.