함수의 극한과 연속 전체 문제

The function $1/\sqrt{x}$ is continuous on $(0,+\infty)$ and the function $\cos(x) + 1$ is continuous everywhere.

So $1/\sqrt{\cos(x) + 1 }$ is continuous except when $\cos x=-1$. This happens when $x$ is an odd multiple of $\pi$. Hence the function is continuous except at $x=\pm \pi, \pm 3\pi, \pm 5\pi, \dots$.

The function is continuous when $\sin(x)\neq 0$. That is, when $x$ is not an integer multiple of $\pi$.

The function is continuous for $x\ne c$ since each of those two branches are polynomials. So, the only question is whether the function is continuous at $x=c$; for this we need $$\lim_{x\to c^-}f(x)=f(c)=\lim_{x\to c+}f(x).$$ We compute $$\lim_{x\to c^-}f(x)=\lim_{x\to c^-}8-cx = 8-c^2;$$ $$f(c)=8-c\cdot c= 8-c^2\text{ and}$$ $$\lim_{x\to c^+}f(x)=\lim_{x\to c^+}x^2=c^2.$$ So, we need $8-c^2=c^2$, which yields $c^2=4$, i.e. $c=-2$ or $c=2$.

The function is continuous for $x \ne 0$ since $x^2+c$ and $\cos cx$ are continuous everywhere. It remains to check continuity at $x=0$. To do this we must check that the following three are equal.

$$\begin{aligned} \lim_{x \to 0^+} f(x) &= \lim_{x\to 0^+} x^2+c = c\\ f(0) &= c \\ \lim_{x \to 0^-} f(x) &= \lim_{x\to 0^-} \cos cx = \cos 0 = 1 \end{aligned}$$

Hence when $c=1$ we have the limits agree.

The function is continuous for $x\ne c$ since each of those two branches are defined by polynomials. Thus, the only question is whether the function is continuous at $x=c$. Furthermore, $$\lim_{x\to c^-}f(x) = c^2-4$$ and $$\lim_{x\to c^+}f(x) = f(c) = 3c\,.$$ For continunity we need both limits and the value to agree, so $f$ is continuous if and only if $c^2-4 = 3c$, that is if and only if $$c^2-3c-4 = 0\,.$$ Factoring this as $(c-4)(c+1) = 0$ yields $c=-1$ or $c=+4$.

The function is continuous for $x\ne 2c$ since each of those two branches are polynomials. So, the only question is whether the function is continuous at $x=2c$; for this we need $$\lim_{x\to 2c^-}f(x)=f(2c)=\lim_{x\to 2c+}f(x).$$ We compute $$\lim_{x\to 2c^-}f(x)=\lim_{x\to 2c^-}6-cx = 6-2c^2;$$ $$f(2c)=6-c\cdot 2c= 6-2c^2\text{ and}$$ $$\lim_{x\to 2c^+}f(x)=\lim_{x\to 2c^+}x^2=4c^2.$$ So, we need $6-2c^2=4c^2$, which yields $c^2=1$, i.e. $c=-1$ or $c=1$.

This isn't the kind of equality that we can just solve; we'll need a trick, and that trick is the IVT. The general idea is to show that $\sin x$ is somewhere bigger, and somewhere smaller, than $x-1$. However, since the IVT can only show us that a function is equal to a constant, we need to slightly adjust our language. Showing $\sin x = x-1$ is equivalent to showing $\sin x - x + 1 = 0$, so let $f(x)=\sin x - x +1$, and let's show that it has a real root.

First, we need to note that $f(x)$ is continuous (otherwise we can't use the IVT). Now, we need to find a value of $x$ for which it is positive, and for which it's negative. By checking a few values, we find $f(0)$ is positive, and $f(100)$ is negative. So, by the IVT, there exists a value of $x$ (between $0$ and $100$) for which $f(x)=0$. Therefore, there exists a value of $x$ for which $\sin x = x-1$.

We let $f(x)=3^x-x^2$. Then $f(x)$ is a continuous function, since both $3^x$ and $x^2$ are continuous for all real numbers.

We find a value $a$ such that $f(a)>0$. We observe immediately that $a=0$ works since $$f(0)=3^0-0=1>0.$$

We find a value $b$ such that $f(b)<0$. We see that $b=-1$ works since $$f(-1)=\frac{1}{3}-1<0.$$

So, because $f(x)$ is continuous on $(-\infty, \infty)$ and $f(0)>0$ while $f(-1)<0$, then the Intermediate Value Theorem guarantees the existence of a real number $c$ in the interval $(-1,0)$ such that $f(c)=0$.

We let $f(x)=2\tan(x)-x-1$. Then $f(x)$ is a continuous function on the interval $(-\pi/2, \pi/2)$ since $\tan(x)=\sin(x)/\cos(x)$ is continuous on this interval, while $x+1$ is a polynomial and therefore continuous for all real numbers.

We find a value $a\in (-\pi/2,\pi/2)$ such that $f(a)<0$. We observe immediately that $a=0$ works since $$f(0)=2\tan(0)-0-1=0-1=-1<0.$$

We find a value $b\in (-\pi/2, \pi/2)$ such that $f(b)>0$. We see that $b=\pi/4$ works since $$f(\pi/4)=2\tan(\pi/4) -\pi/4 - 1=2-\pi/4 - 1=1-\pi/4=(4-\pi)/4>0,$$ because $3<\pi<4$.

So, because $f(x)$ is continuous on $[0,\pi/4]$ and $f(0)<0$ while $f(\pi/4)>0$, then the Intermediate Value Theorem guarantees the existence of a real number $c\in (0,\pi/4)$ such that $f(c)=0$.

Let $f(x) = \sqrt{\cos(\pi x)} - \sin(2\pi x) -1/2$. This function is continuous provided $\cos(\pi x)\geq 0$. This is true for $0 \leq x \leq \frac{1}{2}$.

Now $f$ takes positive values on $[0,1/2]$:

$$\begin{aligned} f(0) &= \sqrt{\cos(0)} - \sin(0) -1/2 = \sqrt{1} -1/2 = 1/2. \end{aligned}$$

And $f$ takes negative values on $[0,1/2]$:

$$\begin{aligned} f(1/2) &= \sqrt{\cos(\pi/2)}-\sin(\pi)-1/2 = 0-0-1/2 = -1/2 \end{aligned}$$

(Notice that $f(1/3)=(\sqrt{2}-\sqrt{3})/2-1/2$ also works)

So, because $f(x)$ is continuous on $[0,1/2)$ and $f(0)>0$ while $f(1/2)<0$, then the Intermediate Value Theorem guarantees the existence of a real number $c\in (0,1/2)$ such that $f(c)=0$.

We let $f(x)=\dfrac{1}{\cos^2(\pi x)}-x-\dfrac{3}{2}$. Then $f(x)$ is a continuous function on the interval $(-1/2, 1/2)$ since $\cos x$ is continuous everywhere and non-zero on that interval.

The function $f$ takes negative values. For example, when $x=0$:

$$\begin{aligned} f(0) &= \frac{1}{\cos^2(0)} - 0 - \frac{3}{2} = 1-\frac{3}{2} = -\frac{1}{2} <0. \end{aligned}$$

It also takes positive values, for instance when $x=1/4$:

$$\begin{aligned} f(1/4) &= \frac{1}{(\cos \pi/4)^2} - \frac{1}{4} - \frac{3}{2} \\ &= \frac{1}{1/2} - \frac{1+6}{4} \\ &= 2 - 7/4 = 1/4 >0. \end{aligned}$$

By the IVT there is $c$, $0<c<1/4$ such that $f(c)=0$, in which case

$$\begin{aligned} \dfrac{1}{(\cos\pi c)^2} = c+\dfrac{3}{2}. \end{aligned}$$

$f(x)$ is a polynomial, so it's continuous everywhere. If we can find values $a$ and $b$ so that $f(a)>0$ and $f(b)<0$, then by the IVT, there will exist some $c$ in $(a,b)$ where $f(c)=0$; that is, there is a root in the interval $[a,b]$. Let's start plugging in easy values of $x$.

$f(0)=15$, and $f(1)=1-15+9-18+15=-8$. Since 0 is between $f(0)$ and $f(1)$, and since $f$ is continuous, by IVT there must be some $x$ in $[0,1]$ for which $f(x)=0$: that is, there is some root in $[0,1]$.

That was the easiest interval to find. If you keep playing around, you find two more. $f(-1)=26$ (positive) and $f(-2)=-1001$ (negative), so there is a root in $[-2,-1]$.

The arithmetic is nasty, but there is also a root in $[14,15]$.

This is an important trick. For high-degree polynomials, it is often impossible to get the exact values of the roots. Using the IVT, we can at least give a range where a root must be.

Let $f(x)=x^3$. Since $f$ is a polynomial, it is continuous everywhere. If $f(a)<7<f(b)$, then $\sqrt[3]{7}$ is somewhere between $a$ and $b$. If we can find $a$ and $b$ that satisfy these inequalities, and are very close together, that will give us a good approximation for $\sqrt[3]{7}$.

• Let's start with integers. $1^3<7<2^3$, so $\sqrt[3]{7}$ is in the interval $(1,2)$.
• Let's narrow this down, say by testing $f(1.5)$. $(1.5)^3 = 3.375<7$, so $\sqrt[3]{7}$ is in the interval $(1.5,2)$.
• Let's narrow further, say by testing $f(1.75)$. $(1.75)^3\approx 5.34<7$, so $\sqrt[3]{7}$ is in the interval $(1.75,2)$.
• Testing various points, we find $f(1.9)<7<f(2)$, so $\sqrt[3]{7}$ is between $1.9$ and $2$.
• By testing more, we find $f(1.91)<7<f(1.92)$, so $\sqrt[3]{7}$ is in $(1.91,1.92)$.
• In order to get an approximation for $\sqrt[3]{7}$ that is rounded to two decimal places, we have to know whether $\sqrt[3]{7}$ is greater or less than $1.915$; indeed $f(1.915) \approx 7.02>7$, so $\sqrt[3]{7}<1.915$; then rounded to two decimal places, $\sqrt[3]{7}\approx 1.91$.

If this seems like the obvious way to approximate $\sqrt[3]{7}$, that's good. The IVT is a formal statement of a very intuitive principle.

• If $f(a)=g(a)$, or $f(b)=g(b)$, then we simply take $c=a$ or $c=b$.
• Suppose $f(a) \neq g(a)$ and $f(b) \neq g(b)$. Then $f(a)<g(a)$ and $g(b)<f(b)$, so if we define $h(x)=f(x)-g(x)$, then $h(a)<0$ and $h(b)>0$. Since $h$ is the difference of two functions that are continuous over $[a,b]$, also $h$ is continuous over $[a,b]$. So, by the Intermediate Value Theorem, there exists some $c \in (a,b)$ with $h(c)=0$; that is, $f(c)=g(c)$.