함수의 극한과 연속 전체 문제

We divide both the numerator and the denominator by the highest power of $x$ in the denominator, which is $x$. Since $x<0$, we have $\sqrt{x^2}=|x|=-x$, so that $$\frac{\sqrt{4x^2+15}}{x}= \frac{\sqrt{4x^2+15}}{-\sqrt{x^2}} =-\sqrt{\frac{4x^2+15}{x^2}}=-\sqrt{4+\frac{15}{x^2}}.$$ Since $1/x\to 0$ and also $1/x^2\to 0$ as $x\to -\infty$, we conclude that $$\lim_{x\to -\infty} \frac{5x+7}{\sqrt{4x^2+15}-x}=\lim_{x\to -\infty}\frac{5+\frac{7}{x}}{-\sqrt{4+\frac{15}{x^2}}-1}=\frac{5}{-2-1}=-\frac{5}{3}.$$

$$\begin{aligned} \displaystyle\lim_{x \rightarrow -\infty}\dfrac{3x^7+x^5-15}{4x^2+32x}&= \displaystyle\lim_{x \rightarrow -\infty}\dfrac{x^2(3x^5+x^3-\frac{15}{x^2})}{x^2(4+\frac{32}{x})}\\ &=\displaystyle\lim_{x \rightarrow -\infty}\dfrac{3x^5+x^3-\frac{15}{x^2}}{4+\frac{32}{x}} \\ &=\displaystyle\lim_{x \rightarrow +\infty}\dfrac{3(-x)^5+(-x)^3-\frac{15}{(-x)^2}}{4+\frac{32}{-x}}\\ &=\displaystyle\lim_{x \rightarrow +\infty}\dfrac{-3x^5-x^3-\frac{15}{x^2}}{4-\frac{32}{x}} \\&=-\infty \end{aligned}$$

We multiply and divide the expression by its conjugate, $\big(\sqrt{n^2+5n}+n\big)$.

$$\begin{aligned} \lim_{n\rightarrow\infty}\big(\sqrt{n^2+5n}-n\big) &=\lim_{n \to \infty}\big(\sqrt{n^2+5n}-n\big) \left(\frac{\sqrt{n^2+5n}+n}{\sqrt{n^2+5n}+n}\right)\\ &=\lim_{n\rightarrow\infty}\frac{(n^2+5n)-n^2}{\sqrt{n^2+5n}+n}\\ &=\lim_{n\rightarrow\infty}\frac{5n}{\sqrt{n^2+5n}+n}\\ &=\lim_{n\rightarrow\infty}\frac{5\cdot n}{\sqrt{n^2}\sqrt{1+\frac{5}{n}}+n} \\ \text{Since $n>0$, we can simplify $\sqrt{n^2}=n$.} \\ &=\lim_{n\rightarrow\infty}\frac{5\cdot n}{n\sqrt{1+\frac{5}{n}}+n}\\ &=\lim_{n\rightarrow\infty}\frac{5}{\sqrt{1+\frac{5}{n}}+1}\\ &=\frac{5}{\sqrt{1+0}+1}=\frac{5}{2} \end{aligned}$$

Since $x=3$ is not in the domain of the function, we simplify, hoping we can cancel a problematic term.

$$\begin{aligned} \lim_{x \to 3}\frac{2x+8}{\frac{1}{x-3}+\frac{1}{x^2-9}}&=\lim_{x \to 3}\frac{2x+8}{\frac{x+3}{x^2-9}+\frac{1}{x^2-9}}\\ &=\lim_{x \to 3}\frac{2x+8}{\frac{x+4}{x^2-9}}\\ &=\lim_{x \to 3}\frac{(2x+8)(x^2-9)}{x+4}=0 \end{aligned}$$

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.

Yes. This is a straightforward application of IVT.

No. IVT says that $f(x)=0$ for some $x$ between $10$ and $20$, but it doesn't have to be exactly half way between.

No. IVT says nothing about functions that are not guaranteed to be continuous at the outset. It's quite easy to construct a function that is as described, but not continuous. For example, the function pictured below, whose equation format is somewhat less enlightening than its graph: $f(x)=\left\{\begin{array}{ll} -\frac{26}{5}x+65,&10 \leq x < 15\\ -\frac{26}{5}x+91,&15 \leq x \leq 20 \end{array}\right.$.

Since $f(x)$ and $g(x)$ are continuous at zero, and since $g^2(x)+1$ must be nonzero, then $h(x)$ is continuous at 0 as well. According to the definition of continuity, then $\displaystyle \lim_{x \to 0}h(x)$ exists and is equal to $h(0)=\frac{0f(0)}{g^2(0)+1}=0$.

Since the limit $\displaystyle \lim_{x \to 0}h(x)$ exists and is equal to zero, also the one-sided limit $\displaystyle \lim_{x \to 0^+}h(x)$ exists and is equal to zero.

$f(x)$ is a rational function and so is continuous except when its denominator is zero. That is, except when $x=1$ and $x=-1$.

The function is continuous when $x^2-1> 0$, i.e. $(x-1)(x+1)> 0$, which yields the intervals $(-\infty, -1)\cup (1,+\infty)$.

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 $\pi$.

$f(x)$ 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.