함수의 극한과 연속 전체 문제

(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.

$f(x)$ 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.