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 $$\frac{\sqrt{x^2+5}}{x}=-\sqrt{\frac{x^2+5}{x^2}}=-\sqrt{1+\frac{5}{x^2}}.$$ Since $1/x\to 0$ and also $1/x^2\to 0$ as $x\to -\infty$, we conclude that $$\lim_{x\to -\infty} \frac{3x+5}{\sqrt{x^2+5}-x}=\lim_{x\to -\infty}\frac{3+\frac{5}{x}}{-\sqrt{1+\frac{5}{x^2}}-1}=\frac{3}{-1-1}=-\frac{3}{2}.$$

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.

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.

Many answers are possible; the tangent function behaves like this.

If we let $f$ be my height, the time of my birth is $a$, and now is $b$, then we know that $f(a) \leq 1 \leq f(b)$. It is reasonable to assume that my height is a continuous function. So by the IVT, there is some value $c$ between $a$ and $b$ where $f(c)=1$. That is, there is some time (we called it $c$) between my birth and today when I was exactly one meter tall.

Notice the IVT does not say precisely what day I was one meter tall; it only guarantees that such a day occurred between my birth and today.

One example is $f(x) = \left\{ \begin{array}{ll} 0&\text{when }0 \leq x \leq 1\\ 2&\text{when }1<x \leq 2 \end{array}\right.$. The IVT only guarantees $f(c)=1$ for some $c$ in $[0,2]$ when $f$ is continuous over $[0,2]$. If $f$ is not continuous, the IVT says nothing.

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: $f(x)=\left\{\begin{array}{ll} -\frac{26}{5}x+65,&10 \leq x < 15\\ -\frac{26}{5}x+91,&15 \leq x \leq 20 \end{array}\right.$.

True. Since $f(t)$ is continuous at $t=5$, that means $\displaystyle \lim_{t \to 5} f(t)=f(5)$. For that to be true, $f(5)$ must exist — that is, 5 is in the domain of $f(x)$.

True. Using the definition of continuity, $\displaystyle \lim_{t \to 5} f(t)=f(5)=17$.

In general, false. If $f(t)$ is continuous at $t=5$, then $f(5)=17$; if $f(t)$ is discontinuous at $t=5$, then $f(5)$ either does not exist, or is a number other than 17.

An example of a function with $\displaystyle \lim_{t \to 5}f(t)=17 \neq f(5)$ is $f(t)=\left\{ \begin{array}{lcr} 17&,&t \neq 5\\ 0&,&t=5 \end{array} \right.$, shown below.

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.

Since $f$ is a polynomial, it is continuous over all real numbers. $f(0)=1<12345$ and $f(12345)=12345^3+12345^2+12345+1>12345$ (since all terms are positive). So by the IVT, $f(c)=12345$ for some $c$ between $0$ and $12345$.

$f(x)$ 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 $(-\infty, -1)\cup (1,+\infty)$.