극한의 성질과 계산

$-2$

$-3/2$

$-7/8$

$0$

$0$

$1$

$9$

$3$

$5/8$

$1$

$\pi/180$

The statement $\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=10$ tells us that, as $x$ gets very close to 3, $f(x)$ is 10 times as large as $g(x)$. We notice that if $f(x)=10g(x)$, then $\dfrac{f(x)}{g(x)}=10$, so $\displaystyle\lim_{x \rightarrow} \dfrac{f(x)}{g(x)}=10$ wherever $f$ and $g$ exist. So it's enough to find a function $g(x)$ that has limit 0 at 3. Such a function is (for example) $g(x)=x-3$. So, we take $f(x)=10(x-3)$ and $g(x)=x-3$. It is easy now to check that $\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$ and $\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=\displaystyle \lim_{x \to 3}\frac{10(x-3)}{x-3}=\displaystyle \lim_{x \to 3}10=10$.

One way to start this problem is to remember $\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x^2}=\infty$. (Using $\dfrac{1}{x^2}$ as opposed to $\dfrac{1}{x}$ is important, since $\displaystyle\lim_{x \rightarrow 0}\dfrac{1}{x}$ does not exist.) Then by “shifting" by three, we find $\displaystyle \lim_{x \rightarrow 3} \dfrac{1}{(x-3)^2}=\infty$. So it is enough to arrange that $\dfrac{f(x)}{g(x)}=\dfrac{1}{(x-3)^2}$. We can achieve this with $f(x)=x-3$ and $g(x)=(x-3)^3$, and maintain $\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$. Again, this is one of many possible solutions.

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 $\displaystyle \lim_{x \to -4}f(x)=0$.