함수의 극한과 연속 전체 문제

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 $x$, $\sqrt{x^2}=|x|=-x$.

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

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

$$\begin{aligned} \displaystyle\lim_{x \rightarrow \infty}\frac{x+1}{\sqrt{x^2}}&= \displaystyle\lim_{x \rightarrow \infty}\frac{x+1}{x}\\ &=\displaystyle\lim_{x \rightarrow \infty}1+\frac{1}{x}=1+0=1 \end{aligned}$$

When $x<0$, $|x|=-x$ and so $\displaystyle \lim_{x\to\infty}\sin\left( \frac{\pi}{2} \cdot \frac{|x|}{x}\right) + \frac{1}{x} = \sin(-\pi/2) = -1$.

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, $\big(\sqrt{n^2+5n}+n\big)$.

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