함수의 극한과 연속 전체 문제

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

Any polynomial of degree one or higher will go to $\infty$ or $-\infty$ as $x$ goes to $\infty$. So, we need a polynomial of degree 0–that is, $f(x)$ is a constant. One possible answer is $f(x)=1$.

This will be the case for any polynomial of odd degree. For instance, $f(x)=x$.

Many answers are possible: also $f(x)=x^{15}-32x^2+9$ satisfies $\displaystyle\lim_{x \rightarrow \infty} f(x) = \infty$ and $\displaystyle\lim_{x \rightarrow -\infty} f(x)= -\infty$.

$\displaystyle\lim_{x \rightarrow \infty} 2^{-x}= \displaystyle\lim_{x \rightarrow \infty} \frac{1}{2^x}=0$

As $x$ gets larger and larger, $2^x$ grows without bound. (For integer values of $x$, you can imagine multiplying $2$ by itself more and more times.) So, $\displaystyle\lim_{x \rightarrow \infty}2^x=\infty$.

There is no single number that $\cos x$ approaches as $x$ becomes more and more strongly negative: as $x$ grows in the negative direction, the function oscillates between $-1$ and $+1$, never settling close to one particular number. So, this limit does not exist.

The highest-order term in this polynomial is $-3x^5$, so this dominates the function's behaviour as $x$ goes to infinity. More formally:

$$\begin{aligned} \lim\limits_{x\rightarrow\infty} \big(x - 3x^5 +100 x^2\big) &=\lim\limits_{x\rightarrow\infty} - 3x^5\left(1 -\frac{1}{3x^4} -\frac{100}{3 x^3}\right) \\ & =\lim\limits_{x\rightarrow\infty} - 3x^5 =-\infty \\ \text{because} \\ \lim\limits_{x\rightarrow\infty} \left(1 -\frac{1}{3x^4} -\frac{100}{3 x^3}\right) &=1-0-0=1. \end{aligned}$$

Our standard trick is to factor out the highest power of $x$ in the denominator: $x^4$. We just have to be a little careful with the square root. Since we are taking the limit as $x$ goes to positive infinity, we have positive $x$-values, so $\sqrt{x^2}=x$ and $\sqrt{x^8}=x^4$.

$$\begin{aligned} \displaystyle\lim_{x \rightarrow\infty} \dfrac{\sqrt{3x^8+7x^4}+10}{x^4-2x^2+1}&= \displaystyle\lim_{x \rightarrow\infty} \dfrac{\sqrt{x^8(3+\frac{7}{x^4})}+10}{x^4(1-\frac{2}{x^2}+\frac{1}{x^4})} \\&= \displaystyle\lim_{x \rightarrow\infty} \dfrac{\sqrt{x^8}\sqrt{3+\frac{7}{x^4}}+10}{x^4(1-\frac{2}{x^2}+\frac{1}{x^4})}\\&= \displaystyle\lim_{x \rightarrow\infty} \dfrac{x^4\sqrt{3+\frac{7}{x^4}}+10}{x^4(1-\frac{2}{x^2}+\frac{1}{x^4})}\\&= \displaystyle\lim_{x \rightarrow\infty} \dfrac{x^4\left(\sqrt{3+\frac{7}{x^4}}+\frac{10}{x^4}\right)}{x^4(1-\frac{2}{x^2}+\frac{1}{x^4})}\\&= \displaystyle\lim_{x \rightarrow\infty} \dfrac{\sqrt{3+\frac{7}{x^4}}+\frac{10}{x^4}}{1-\frac{2}{x^2}+\frac{1}{x^4}}\\ &=\frac{\sqrt{3+0}+0}{1-0+0}=\sqrt{3} \end{aligned}$$

We have two terms, each getting extremely large. It's unclear at first what happens when we subtract them. To get this equation into another form, we multiply and divide by the conjugate, $\sqrt{x^2+5x}+\sqrt{x^2-x}$.

$$\begin{aligned} \lim_{x\rightarrow \infty}\left[\sqrt{x^2+5x}-\sqrt{x^2-x}\right] &=\lim_{x\rightarrow \infty}\left[\dfrac{(\sqrt{x^2+5x}-\sqrt{x^2-x})(\sqrt{x^2+5x}+\sqrt{x^2-x})}{\sqrt{x^2+5x}+\sqrt{x^2-x}}\right]\\ &= \lim_{x\rightarrow \infty}\dfrac{(x^2+5x)-(x^2-x)} {\sqrt{x^2+5x}+\sqrt{x^2-x}}\\ & = \lim_{x\rightarrow \infty}\dfrac{6x}{\sqrt{x^2+5x}+\sqrt{x^2-x}} \\ \text{Now we divide the numerator and denominator by $x$. In the case of the denominator, since $x>0$, $x=\sqrt{x^2}$.} \\ &= \lim_{x\rightarrow \infty}\dfrac{6(x)}{\sqrt{x^2}\sqrt{1+\frac{5}{x}}+\sqrt{x^2}\sqrt{1-\frac{1}{x}}} \\&= \lim_{x\rightarrow \infty}\dfrac{6(x)}{(x)\sqrt{1+\frac{5}{x}}+(x)\sqrt{1-\frac{1}{x}}} \\& = \lim_{x\rightarrow \infty}\dfrac{6}{\sqrt{1+\frac{5}{x}}+\sqrt{1-\frac{1}{x}}}\\ &=\frac{6}{\sqrt{1+0}+\sqrt{1-0}}= 3 \end{aligned}$$

Note that for large negative $x$, the first term in the denominator $\sqrt{4x^2+x}\approx\sqrt{4x^2}=|2x|=-2x$ not $+2x$. A good way to avoid incorrectly computing $\sqrt{x^2}$ when $x$ is negative is to define $y=-x$ and express everything in terms of $y$. That's what we'll do.

$$\begin{aligned} \lim_{x\to -\infty} \frac{3x}{\sqrt{4x^2+x}-2x} &=\lim_{y\to +\infty} \frac{-3y}{\sqrt{4y^2-y}+2y} \\ &=\lim_{y\to +\infty} \frac{-3y}{y\sqrt{4-\frac{1}{y}}+2y} \\ &=\lim_{y\to +\infty} \frac{-3}{\sqrt{4-\frac{1}{y}}+2} \\ &=\frac{-3}{\sqrt{4-0}+2} \quad\text{since $1/y\to 0$ as $y\to +\infty$} \\ &= -\frac{3}{4} \end{aligned}$$

The highest power of $x$ in the denominator is $x^2$, so we divide the numerator and denominator by $x^2$:

$$\begin{aligned} \lim\limits_{x\rightarrow -\infty}\frac{1-x-x^2}{2x^2-7}&= \lim\limits_{x\rightarrow -\infty}\frac{1/x^2-1/x-1}{2-7/x^2}\\ &=\frac{0-0-1}{2-0}=-\frac{1}{2} \end{aligned}$$

$$\begin{aligned} \lim_{x\rightarrow\infty}\big(\sqrt{x^2+x}-x\big) &=\lim_{x\rightarrow\infty} \frac{\big(\sqrt{x^2+x}-x\big)\big(\sqrt{x^2+x}+x\big)} {\sqrt{x^2+x}+x} =\lim_{x \to \infty}\frac{(x^2+x)-x^2}{\sqrt{x^2+x}+x}\\ &=\lim_{x\rightarrow\infty} \frac{x}{\sqrt{x^2+x}+x} =\lim_{x\rightarrow\infty} \frac{1}{\sqrt{1+\frac{1}{x}}+1} =\frac{1}{2} \end{aligned}$$

We have, after dividing both numerator and denominator by $x^2$ (which is the highest power of the denominator) that $$\frac{5x^2-3x+1}{3x^2+x+7}=\frac{5-\frac{3}{x}+\frac{1}{x^2}}{3+\frac{1}{x}+\frac{7}{x^2}}.$$ Since $1/x\to 0$ and also $1/x^2\to 0$ as $x\to +\infty$, we conclude that $$\lim_{x\to +\infty} \frac{5x^2-3x+1}{3x^2 +x+7}=\frac{5}{3}.$$

We have, after dividing both numerator and denominator by $x$ (which is the highest power of the denominator) that $$\frac{ \sqrt{4\,x + 2}}{3x+4}=\frac{\sqrt{\frac 4 x + \frac 2 {x^2}}}{3 + \frac 4 x}.$$ Since $1/x\to 0$ and also $1/x^2\to 0$ as $x\to +\infty$, we conclude that $$\lim_{x\to +\infty} \frac{ \sqrt{4\,x + 2}} {3\,x+4}=\frac {0}{3} = 0.$$

The dominant terms in the numerator and denominator have order $x^3$. Taking out that common factor we get $$\frac{4x^3+x}{7x^3 + x^2 - 2} = \frac{4 + \frac{1}{x^2}}{7 + \frac{1}{x} - \frac{2}{x^3}}.$$ Since $1/x^a\to 0$ as $x\to +\infty$ (for $a>0$), we conclude that $$\lim_{x\to +\infty} \frac{4x^3+x}{7x^3 +x^2-2}=\frac{4}{7}.$$

We have, after dividing both numerator and denominator by $x^3$ (which is the highest power of the denominator) that: $$\lim_{x \to \infty} \frac{5x^2+10}{3x^3 +2x^2+x}=\lim_{x \to\infty}\frac{\frac{5}{x}+\frac{10}{x^3}}{3+\frac{2}{x}+\frac{1}{x^2}}=\frac{0}{3}=0.$$

Since we only consider negative values of $x$, $\sqrt{x^2}=|x|=-x$.

$$\begin{aligned} \displaystyle\lim_{x \rightarrow -\infty}\frac{x+1}{\sqrt{x^2}}&= \displaystyle\lim_{x \rightarrow -\infty}\frac{x+1}{-x}\\ &=\displaystyle\lim_{x \rightarrow -\infty}\frac{x}{-x}+\frac{1}{-x}\\ &=\displaystyle\lim_{x \rightarrow -\infty}-1-\frac{1}{x}\\ &=-1 \end{aligned}$$