극한·미분 심화 전체 문제

$-0.521095$

$10$

$\epsilon$ should be given first, and the restriction $|x-a|<\delta$ implies $|f(x)-K|< \epsilon$, not the other way around.

The $y$–tolerance.

T

T

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-4|<\delta$, $|f(x)-13|<\epsilon$.

Consider $|f(x)-13|<\epsilon$:

$$\begin{gathered} |f(x) -13 | < \epsilon \\ |(2x+5)-13 |<\epsilon \\ | 2x-8 | < \epsilon \\ 2| x-4 | < \epsilon \\ -\epsilon/2 < x-4 < \epsilon/2. \\ \end{gathered}$$

This implies we can let $\delta =\epsilon/2$. Then:

$$\begin{gathered} |x-4|<\delta \\ -\delta < x-4 < \delta\\ -\epsilon/2 < x-4 < \epsilon/2\\ -\epsilon < 2x-8 < \epsilon \\ -\epsilon < (2x+5)-13 < \epsilon \\ |(2x+5) - 13| < \epsilon, \end{gathered}$$

which is what we wanted to prove.

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-5|<\delta$, $|f(x)-(-2)|<\epsilon$.

Consider $|f(x)-(-2)|<\epsilon$:

$$\begin{gathered} |f(x) + 2 | < \epsilon \\ |(3-x) + 2 |<\epsilon \\ | 5-x | < \epsilon \\ -\epsilon < 5-x < \epsilon \\ -\epsilon < x-5 < \epsilon. \\ \end{gathered}$$

This implies we can let $\delta =\epsilon$. Then:

$$\begin{gathered} |x-5|<\delta \\ -\delta < x-5 < \delta\\ -\epsilon < x-5 < \epsilon\\ -\epsilon < (x-3)-2 < \epsilon \\ -\epsilon < (-x+3)-(-2) < \epsilon \\ |3-x - (-2)| < \epsilon, \end{gathered}$$

which is what we wanted to prove.

Let $\epsilon >0$ be given. We wish to find $\delta >0$ such that when $|x-3|<\delta$, $|f(x)-6|<\epsilon$.

Consider $|f(x)-6|<\epsilon$, keeping in mind we want to make a statement about $|x-3|$:

$$\begin{gathered} |f(x) -6 | < \epsilon \\ |x^2-3 -6 |<\epsilon \\ | x^2-9 | < \epsilon \\ | x-3 |\cdot|x+3| < \epsilon \\ | x-3 | < \epsilon/|x+3| \\ \end{gathered}$$

Since $x$ is near 3, we can safely assume that, for instance, $2<x<4$. Thus

$$\begin{gathered} 2+3<x+3<4+3 \\ 5 < x+3 < 7 \\ \frac{1}{7} < \frac{1}{x+3} < \frac{1}{5} \\ \frac{\epsilon}{7} < \frac{\epsilon}{x+3} < \frac{\epsilon}{5} \\ \end{gathered}$$

Let $\delta =\frac{\epsilon}{7}$. Then:

$$\begin{gathered} |x-3|<\delta \\ |x-3| < \frac{\epsilon}7\\ |x-3| < \frac{\epsilon}{x+3}\\ |x-3|\cdot|x+3| < \frac{\epsilon}{x+3}\cdot|x+3|\\ \end{gathered}$$

Assuming $x$ is near 3, $x+3$ is positive and we can drop the absolute value signs on the right.

$$\begin{gathered} |x-3|\cdot|x+3| < \frac{\epsilon}{x+3}\cdot(x+3)\\ |x^2-9| < \epsilon\\ |(x^2-3) - 6| < \epsilon, \end{gathered}$$

which is what we wanted to prove.

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

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.