01. 극한의 엄밀한 정의Precise Definition of a Limit

Solution. Since $y = 2x-1$, we have $y-7 = 2x-8 = 2(x-4)$.

(1) $|y-7| \lt 2 \Leftrightarrow |2(x-4)| \lt 2 \Leftrightarrow |x-4| \lt 1$, so the range of $x$ is $3 \lt x \lt 5$.

(2) $|y-7| \lt 1 \Leftrightarrow |2(x-4)| \lt 1 \Leftrightarrow |x-4| \lt \dfrac{1}{2}$, so the range of $x$ is $3.5 \lt x \lt 4.5$.

(3) $|y-7| \lt \epsilon \Leftrightarrow |2(x-4)| \lt \epsilon \Leftrightarrow |x-4| \lt \dfrac{\epsilon}{2}$, so the range of $x$ is $4-\dfrac{\epsilon}{2} \lt x \lt 4+\dfrac{\epsilon}{2}$.

Solution. $y = \sqrt{x}$.

(1) $|y-2| \lt 1 \Leftrightarrow 1 \lt \sqrt{x} \lt 3 \Leftrightarrow 1 \lt x \lt 9$.

(2) By (1), the range of $x$ is $1 \lt x \lt 9$, so it suffices to have $4-\delta \geq 1$ and $4+\delta \leq 9$ whenever $|x-4| \lt \delta$, i.e., $\delta \leq 3$ and $\delta \leq 5$. Taking $\delta = 3$ gives $|x-4| \lt 3 \Rightarrow 1 \lt x \lt 7 \subset (1,9) \Rightarrow |y-2| \lt 1$.

(3) $|y-1| \lt 0.5 \Leftrightarrow 0.5 \lt \sqrt{x} \lt 1.5 \Leftrightarrow 0.25 \lt x \lt 2.25$.

(4) By (3), the range of $x$ is $0.25 \lt x \lt 2.25$, so it suffices to have $1-\delta \geq 0.25$ and $1+\delta \leq 2.25$, i.e., $\delta \leq 0.75$ and $\delta \leq 1.25$. Taking $\delta = 0.75$ gives $|x-1| \lt 0.75 \Rightarrow 0.25 \lt x \lt 1.75 \subset (0.25, 2.25) \Rightarrow |y-1| \lt 0.5$.

Solution. $y = mx+b$ with $m \gt 0$.

Since $y - \left(\dfrac{m}{2}+b\right) = mx+b-\dfrac{m}{2}-b = m\left(x-\dfrac{1}{2}\right)$, if $|x-\frac12| \lt \delta$ then $\left|y-\left(\dfrac{m}{2}+b\right)\right| = m\left|x-\dfrac{1}{2}\right| \lt m\delta$.

Hence taking $\delta = \dfrac{1}{2m}$ (so that $m\delta \leq \dfrac{1}{2}$) gives $|x-\frac12| \lt \delta \Rightarrow \left|y-\left(\dfrac{m}{2}+b\right)\right| \lt \dfrac{1}{2}$.

Proof.

(1) Let $\epsilon \gt 0$ be given. Take $\delta = \dfrac{\epsilon}{5}$. If $0 \lt |x-1| \lt \delta$, then $|(5x-3)-2| = |5x-5| = 5|x-1| \lt 5\delta = \epsilon$.

(2) Let $\epsilon \gt 0$ be given. Take $\delta = \epsilon$. If $0 \lt |x-c| \lt \delta$, then $|x-c| \lt \delta = \epsilon$.

(3) Let $\epsilon \gt 0$ be given. Take $\delta = 1$ (since $k$ is constant, any $\delta$ works). If $0 \lt |x-c| \lt \delta$, then $|k-k| = 0 \lt \epsilon$.

Proof.

(1) For $x \neq 2$, $\dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2$, which suggests $\displaystyle\lim_{x \to 2} \frac{x^2-4}{x-2} = 4$. Let $\epsilon \gt 0$ be given. Take $\delta = \epsilon$. If $0 \lt |x-2| \lt \delta$, then $\left|\dfrac{x^2-4}{x-2}-4\right| = |(x+2)-4| = |x-2| \lt \delta = \epsilon$.

(2) For $x \neq -5$, $x^2+6x+5 = (x+5)(x+1)$, so $\dfrac{x^2+6x+5}{x+5} = x+1$, which suggests $\displaystyle\lim_{x \to -5} \frac{x^2+6x+5}{x+5} = -4$. Let $\epsilon \gt 0$ be given. Take $\delta = \epsilon$. If $0 \lt |x-(-5)| \lt \delta$, then $\left|\dfrac{x^2+6x+5}{x+5}-(-4)\right| = |(x+1)+4| = |x+5| \lt \delta = \epsilon$.

Proof.

(1) Let $\epsilon \gt 0$ be given. Take $\delta = \min(1, 3\epsilon)$. If $0 \lt |x-5| \lt \delta \leq 1$, then $4 \lt x \lt 6$, so $3 \lt x-1 \lt 5$, hence $\sqrt{x-1} \gt \sqrt{3} \gt 1$ and $\sqrt{x-1}+2 \gt 3$. Then

$$|\sqrt{x-1}-2| = \frac{|(x-1)-4|}{\sqrt{x-1}+2} = \frac{|x-5|}{\sqrt{x-1}+2} \lt \frac{|x-5|}{3} \lt \frac{\delta}{3} \leq \epsilon.$$

(2) Since $f(x) = x^2$ for $x \neq 2$ and $f(2)=1$, for $x$ with $0 \lt |x-2| \lt \delta$ we have $f(x) = x^2$. Let $\epsilon \gt 0$ be given. Take $\delta = \min\left(1, \dfrac{\epsilon}{5}\right)$. If $0 \lt |x-2| \lt \delta \leq 1$, then $1 \lt x \lt 3$, so $|x+2| \lt 5$, hence

$$|f(x)-4| = |x^2-4| = |x-2||x+2| \lt 5\delta \leq \epsilon.$$

Proof.

(1) Let $\epsilon \gt 0$ be given. Take $\delta = \min\left(\dfrac{\sqrt{3}}{2}, \dfrac{3\sqrt{3}}{10}\epsilon\right)$. If $0 \lt |x-\sqrt{3}| \lt \delta \leq \dfrac{\sqrt{3}}{2}$, then $\dfrac{\sqrt{3}}{2} \lt x \lt \dfrac{3\sqrt{3}}{2}$, so $x^2 \gt \dfrac{3}{4}$, i.e., $3x^2 \gt \dfrac{9}{4}$, and $x+\sqrt{3} \lt \dfrac{5\sqrt{3}}{2}$. Hence

$$\left|\frac{1}{x^2}-\frac{1}{3}\right| = \frac{|x-\sqrt3||x+\sqrt3|}{3x^2} \lt \frac{|x-\sqrt3|\cdot\frac{5\sqrt3}{2}}{\frac{9}{4}} = |x-\sqrt3|\cdot\frac{10\sqrt3}{9} \lt \delta \cdot \frac{10\sqrt3}{9} \leq \epsilon.$$

(2) Since $|\sin(1/x)| \leq 1$ for all $x \neq 0$, this suggests $\displaystyle\lim_{x \to 0} x\sin\frac1x = 0$. Let $\epsilon \gt 0$ be given. Take $\delta = \epsilon$. If $0 \lt |x-0| \lt \delta$, then $\left|x\sin\dfrac1x - 0\right| = |x|\left|\sin\dfrac1x\right| \leq |x| \lt \delta = \epsilon$.

(3) Note $f(1) = 6(1)-4 = 2$. For $x \lt 1$, $|f(x)-2| = |(4-2x)-2| = 2|x-1|$, and for $x \geq 1$, $|f(x)-2| = |(6x-4)-2| = 6|x-1|$; in both cases $|f(x)-2| \leq 6|x-1|$. Let $\epsilon \gt 0$ be given. Take $\delta = \dfrac{\epsilon}{6}$. If $0 \lt |x-1| \lt \delta$, then $|f(x)-2| \leq 6|x-1| \lt 6\delta = \epsilon$.

Proof. We argue by negating the definition of the limit: it suffices to find $\epsilon_0 \gt 0$ such that for every $\delta \gt 0$ there exists $x$ with $0 \lt |x-1| \lt \delta$ and $|f(x)-L| \geq \epsilon_0$.

(1) Let $\epsilon_0 = \dfrac12$. For any $\delta \gt 0$, take $x = 1-\dfrac{\delta}{2}$; then $x \lt 1$ and $0 \lt |x-1| = \dfrac{\delta}{2} \lt \delta$. Since $f(x) = x = 1-\dfrac{\delta}{2}$,

$$|f(x)-2| = \left|1-\frac{\delta}{2}-2\right| = 1+\frac{\delta}{2} \gt 1 \gt \frac12 = \epsilon_0.$$

Hence $\displaystyle\lim_{x \to 1} f(x) \neq 2$.

(2) Let $\epsilon_0 = \dfrac12$. For any $\delta \gt 0$, take $x = 1-\dfrac{\delta}{2}$ as in (1); then $x \lt 1$, $0 \lt |x-1| \lt \delta$, and $f(x) = 1-\dfrac{\delta}{2}$, so

$$|f(x)-1.5| = \left|1-\frac{\delta}{2}-1.5\right| = 0.5+\frac{\delta}{2} \gt 0.5 = \epsilon_0.$$

Hence $\displaystyle\lim_{x \to 1} f(x) \neq 1.5$.

Proof. We compute the left-hand and right-hand limits and show they disagree.

(Left-hand limit) We show $\displaystyle\lim_{x \to 2-} h(x) = 4$. Let $\epsilon \gt 0$ be given. Take $\delta = \min\left(1, \dfrac{\epsilon}{4}\right)$. If $2-\delta \lt x \lt 2$, then since $\delta \leq 1$ we have $1 \lt x \lt 2$, so $3 \lt x+2 \lt 4$ and $h(x)=x^2$. Hence

$$|h(x)-4| = |x-2||x+2| \lt 4\delta \leq \epsilon.$$

(Right-hand limit) We show $\displaystyle\lim_{x \to 2+} h(x) = 2$. For $x \gt 2$, $h(x) = 2$, so for any $\epsilon \gt 0$, taking $\delta = 1$ gives $|h(x)-2| = 0 \lt \epsilon$ whenever $2 \lt x \lt 2+\delta$.

For the two-sided limit to exist, the left-hand and right-hand limits must agree; since $4 \neq 2$, $\displaystyle\lim_{x \to 2} h(x)$ does not exist, i.e., it diverges.

Proof.

(1) Let $\epsilon \gt 0$ be given. Take $\delta = \epsilon^2$. If $0 \lt x \lt \delta$, then $\sqrt{x} \lt \sqrt{\delta} = \epsilon$, so $|\sqrt{x}-0| = \sqrt{x} \lt \epsilon$.

(2) For $x \gt 2$, $x-2 \gt 0$, so $|x-2| = x-2$ and hence $\dfrac{x-2}{|x-2|} = 1$. For any $\epsilon \gt 0$, taking $\delta = 1$ gives $\left|\dfrac{x-2}{|x-2|}-1\right| = |1-1| = 0 \lt \epsilon$ whenever $2 \lt x \lt 2+\delta$.

Proof. Note $\lfloor x \rfloor = 400$ for $400 \leq x \lt 401$, and $\lfloor x \rfloor = 399$ for $399 \leq x \lt 400$.

(1) Let $\epsilon \gt 0$ be given. Take $\delta = 1$. If $400 \lt x \lt 400+\delta = 401$, then $\lfloor x \rfloor = 400$, so $|\lfloor x \rfloor - 400| = 0 \lt \epsilon$. Hence $\displaystyle\lim_{x \to 400+} \lfloor x \rfloor = 400$.

(2) Let $\epsilon \gt 0$ be given. Take $\delta = 1$. If $400-\delta = 399 \lt x \lt 400$, then $\lfloor x \rfloor = 399$, so $|\lfloor x \rfloor - 399| = 0 \lt \epsilon$. Hence $\displaystyle\lim_{x \to 400-} \lfloor x \rfloor = 399$.

Solution. By the angle addition formula,

$$\sin\left(\frac{3\pi}{2}+x\right) = \sin\frac{3\pi}{2}\cos x + \cos\frac{3\pi}{2}\sin x = -\cos x,$$

so the given expression is $\cos(2x-\cos x)$. Since $\cos$ and $\sin$ are continuous everywhere and their composition is continuous, the limit equals the function value at $x=\pi/2$. At $x = \pi/2$, $2x = \pi$ and $\cos(\pi/2) = 0$, so

$$\lim_{x \to \pi/2} \cos\left(2x + \sin\left(\frac{3\pi}{2}+x\right)\right) = \cos(\pi - 0) = \cos\pi = -1.$$

Solution.

(1) Since $\tan, \cos, \sin$ are all continuous near $t=0$, the limit of the composition equals its value. As $\tan 0 = 0$ and $\cos 0 = 1$,

$$\lim_{t \to 0} \sin\left(\frac{\pi}{2}\cos(\tan t)\right) = \sin\left(\frac{\pi}{2}\cdot 1\right) = \sin\frac{\pi}{2} = 1.$$

(2) Since $\csc x$ and $\tan x$ are continuous at $x = \pi/6$, the limit equals the function value. As $\sin(\pi/6) = 1/2$, $\csc(\pi/6) = 2$ and $\csc^2(\pi/6) = 4$; and $\tan(\pi/6) = 1/\sqrt3$, so $5\sqrt3\tan(\pi/6) = 5\sqrt3 \cdot \dfrac{1}{\sqrt3} = 5$. Hence

$$\lim_{x \to \pi/6} \sqrt{\csc^2 x + 5\sqrt{3}\tan x} = \sqrt{4+5} = \sqrt9 = 3.$$

Proof.

(1) Let $\epsilon \gt 0$ be given. Take $N = \dfrac{1}{\epsilon}$. If $x \gt N$, then $\left|\dfrac1x - 0\right| = \dfrac1x \lt \dfrac1N = \epsilon$.

(2) Let $\epsilon \gt 0$ be given. Take $N = \dfrac{1}{\epsilon}$. If $x \lt -N$, then $x \lt 0$, so $-x \gt N \gt 0$, and hence $\left|\dfrac1x - 0\right| = \dfrac{1}{-x} \lt \dfrac1N = \epsilon$.

(3) We have $\dfrac{5x^2+8x-3}{3x^2+2} - \dfrac53 = \dfrac{3(5x^2+8x-3)-5(3x^2+2)}{3(3x^2+2)} = \dfrac{24x-19}{3(3x^2+2)}$. For $x \geq 1$, $|24x-19| \leq 24x+19 \leq 24x+19x = 43x$ and $3(3x^2+2) \geq 9x^2$, so

$$\left|\frac{5x^2+8x-3}{3x^2+2} - \frac53\right| \leq \frac{43x}{9x^2} = \frac{43}{9x}.$$

Let $\epsilon \gt 0$ be given. Take $N = \max\left(1, \dfrac{43}{9\epsilon}\right)$. If $x \gt N$, the above inequality holds and $\dfrac{43}{9x} \lt \dfrac{43}{9N} \leq \epsilon$, so $\left|\dfrac{5x^2+8x-3}{3x^2+2} - \dfrac53\right| \lt \epsilon$.

(4) Let $M \gt 0$ be given. Take $\delta = \dfrac{1}{\sqrt{M}}$. If $0 \lt |x| \lt \delta$, then $x^2 \lt \delta^2 = \dfrac1M$, so $\dfrac{1}{x^2} \gt M$.

(5) Let $M \gt 0$ be given. Take $\delta = \min\left(1, \dfrac1M\right)$. If $-2 \lt x \lt -2+\delta$, then since $\delta \leq 1$, $x \in (-2,-1)$, so $0 \lt x+2 \lt \delta$ and $x+3 \in (1,2)$, hence $x+3 \gt 1$. Since $x+2 \gt 0$,

$$\frac{x+3}{x+2} \geq \frac{1}{x+2} \gt \frac{1}{\delta} \geq M.$$

Proof.

(1) Let $\epsilon \gt 0$ be given. Take $N = 1$ (since $k$ is constant, any $N$ works). If $x \gt N$, then $|k-k| = 0 \lt \epsilon$.

(2) Let $M \gt 0$ be given. Take $\delta = \dfrac1M$. If $0 \lt |x| \lt \delta$, then $|x| \lt \dfrac1M$, so $\dfrac{1}{|x|} \gt M$.

Solution. The denominator vanishes at $x=2$, where the numerator is $2^2-3=1 \neq 0$, so $x=2$ is a vertical asymptote.

By polynomial division, $x^2-3 = (2x-4)\left(\dfrac{x}{2}+1\right)+1$, so

$$f(x) = \frac{x}{2}+1+\frac{1}{2x-4}.$$

As $x \to \pm\infty$, $\dfrac{1}{2x-4} \to 0$, so $\displaystyle\lim_{x\to\pm\infty}\left[f(x)-\left(\frac{x}{2}+1\right)\right] = 0$, giving the oblique asymptote $y = \dfrac{x}{2}+1$. (Since the degrees of the numerator and denominator differ by $1$, there is no horizontal asymptote.)

Hence the asymptotes are the vertical asymptote $x=2$ and the oblique asymptote $y = \dfrac{x}{2}+1$.

Solution. The denominator vanishes at $x=0$, where the numerator is $0^3+1=1 \neq 0$, so $x=0$ is a vertical asymptote.

Since $x^3+1 = x^2\cdot x + 1$,

$$f(x) = x+\frac{1}{x^2}.$$

As $x \to \pm\infty$, $\dfrac{1}{x^2} \to 0$, so $\displaystyle\lim_{x\to\pm\infty}[f(x)-x] = 0$, giving the oblique asymptote $y=x$. (Since the degrees of the numerator and denominator differ by $1$, there is no horizontal asymptote.)

Hence the asymptotes are the vertical asymptote $x=0$ and the oblique asymptote $y = x$.