함수의 극한과 연속 전체 문제

We can begin by plotting the points that are easy to read off the diagram. $$\begin{array}{c|c|c} x&f(x)&\frac{1}{f(x)}\\ \hline -3&-3&\frac{-1}{3}\\ -2&0&UND\\ -1&3&\frac{1}{3}\\ 0&3&\frac{1}{3}\\ 1&\frac{3}{2}&\frac{2}{3}\\ 2&0&UND\\ 3&1&1 \end{array}$$ Note that $\frac{1}{f(x)}$ is undefined when $f(x) = 0$. So $\frac{1}{f(x)}$ is undefined at $x=-2$ and $x=2$. We shall look more closely at the behaviour of $\frac{1}{f(x)}$ for $x$ near $\pm 2$ shortly.

Plotting the above points, we get the following picture:

Since $f(x)$ is constant when $x$ is between -1 and 0, then also $\frac{1}{f(x)}$ is constant between -1 and 0, so we update our picture:

The big question that remains is the behaviour of $\frac{1}{f(x)}$ when $x$ is near -2 and 2. We can answer this question with limits. As $x$ approaches $-2$ from the left, $f(x)$ gets closer to zero, and is negative. So $\frac{1}{f(x)}$ will be negative, and will increase in magnitude without bound; that is, $\displaystyle\lim_{x \rightarrow -2^-}\dfrac{1}{f(x)}=-\infty$. Similarly, as $x$ approaches $-2$ from the right, $f(x)$ gets closer to zero, and is positive. So $\frac{1}{f(x)}$ will be positive, and will increase in magnitude without bound; that is, $\displaystyle\lim_{x \rightarrow -2^+}\dfrac{1}{f(x)}=\infty$. We add this behaviour to our graph:

Now, we consider the behaviour at $x=2$. Since $f(x)$ gets closer and closer to 0 AND is positive as $x$ approaches 2, we conclude $\displaystyle\lim_{x \rightarrow 2} \frac{1}{f(x)}=\infty$. Adding to our picture:

Now the only remaining blank space is between $x=0$ and $x=1$. Since $f(x)$ is a smooth curve that stays away from 0, we can draw some kind of smooth curve here, and call it good enough. (Later on we'll go into more details about drawing graphs. The purpose of this exercise was to utilize what we've learned about limits.)

We can start by examining points.

$$\begin{array}{c|c|c|c}x & f(x) & g(x) & \dfrac{f(x)}{g(x)} \\ \hline -3 & -3 & -1.5 & 2 \\ \hline -2 & 0 & 0 & \text{UND} \\ \hline -1 & 3 & 1.5 & 2 \\ \hline -0 & 3 & 1.5 & 2 \\ \hline 1 & 1.5 & .75 & 2 \\ \hline 2 & 0 & 0 & \text{UND} \\ \hline 3 & 1 & .5 & 2\end{array}$$

We cannot divide by zero, so $\dfrac{f(x)}{g(x)}$ is not defined when $x=\pm2$. But for every other value of $x$ that we plotted, $f(x)$ is twice as large as $g(x)$, $\dfrac{f(x)}{g(x)}=2$. With this in mind, we see that the graph of $f(x)$ is exactly the graph of $2g(x)$.

This gives us the graph below.

Remark: $f(2)=g(2)=0$, so $\dfrac{f(2)}{g(2)}$ does not exist, but $\displaystyle\lim_{x \rightarrow 2}\dfrac{f(x)}{g(x)}=2$. Although we are trying to “divide by zero" at $x=\pm 2$, it would be a mistake here to interpret this as a vertical asymptote.

(a) When we evaluate the limit from the left, we only consider values of $x$ that are less than zero. For these values of $x$, our function is $x^2-3$. So, $\displaystyle \lim_{x \to 0^-} f(x)=\displaystyle \lim_{x \to 0^-} (x^2-3)=-3$.

(b) When we evaluate the limit from the right, we only consider values of $x$ that are greater than zero. For these values of $x$, our function is $x^2+3$. So, $\displaystyle \lim_{x \to 0^+} f(x)=\displaystyle \lim_{x \to 0^+} (x^2+3)=3$.

(c) Since the limits from the left and right do not agree, $\displaystyle \lim_{x \to 0} f(x)=$ DNE.

To further clarify the situation, the graph of $y=f(x)$ is sketched below:

(a) When we evaluate $\displaystyle \lim_{x \to -4^-}f(x)$, we only consider values of $x$ that are less than $-4$. For these values, $f(x)=x^3+8x^2+16x$. So, $$\lim_{x \to -4^-}f(x)=\lim_{x \to -4^-} (x^3+8x^2+16x)=(-4)^3+8(-4)^2+16(-4)=0$$ Note that, because $x^3+8x^2+16x$ is a polynomial, we can evaluate the limit by directly substituting in $x=-4$.

(b) When we evaluate $\displaystyle \lim_{x \to -4^+}f(x)$, we only consider values of $x$ that are greater than $-4$. For these values,

$$\begin{aligned} f(x)&=\frac{x^2+8x+16}{x^2+30x-4} \\ \text{So,} \\ \lim_{x \to -4^+}f(x)&=\lim_{x \to -4^+}\frac{x^2+8x+16}{x^2+30x-4} \\ \text{This is a rational function, and $x=-4$ is in its domain (we aren't doing anything suspect, like dividing by 0), so again we can directly substitute $x=-4$ to evaluate the limit:} \\ &=\frac{(-4)^2+8(-4)+16}{(-4)^2+30(-4)-4}=\frac{0}{-108}=0 \end{aligned}$$

(c) Since $\displaystyle \lim_{x\to -4^-}f(x)=\displaystyle \lim_{x \to -4^+}f(x)=0$, we conclude $\displaystyle \lim_{x \to -4}f(x)=0$.

First, we need a rational function whose limit at infinity is a real number. This means that the degree of the bottom is greater than or equal to the degree of the top. There are two cases: the denominator has higher degree than the numerator, or the denominator has the same degree as the numerator.

If the denominator has higher degree than the numerator, then $\displaystyle\lim_{x \rightarrow \infty} f(x)=\displaystyle\lim_{x \rightarrow -\infty} f(x)=0$, so the limits are equal–not what we're looking for.

If the denominator has the same degree as the numerator, then the limit as $x$ goes to $\pm \infty$ is the ratio of the leading terms: again, the limits are equal. So no rational function exists as described.

The amount of the substance that will linger long-term is some positive number–the substance will stick around. One example of a substance that does this is the ink in a tattoo. (If the injection was of medicine, probably it will be metabolized, and $\displaystyle\lim_{t \rightarrow \infty} c(t)=0$.)

Remark: it actually doesn't make much sense to let $t$ go to infinity: after a few million hours, you won't even have a body, so what is $c(t)$ measuring? Often when we use formulas in the real world, there is an understanding that they are only good for some fixed range. We often use the limit as $t$ goes to infinity as a stand-in for the function's long-term behaviour.

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.