함수의 극한과 연속 전체 문제

−∞

$k=\frac{16}{3}$

$\displaystyle\lim_{t \rightarrow 0} \sin t=0$: as $t$ approaches 0, $\sin t$ approaches 0 as well.

$\displaystyle\lim_{x \rightarrow 0^+} \log x = -\infty$: as $x$ approaches 0 from the right, $\log x$ is negative and increasingly large, growing without bound.

$\displaystyle\lim_{y \rightarrow 3} y^2=9$: as $y$ gets closer and closer to 3, $y^2$ gets closer and closer to $3^2$.

$\displaystyle\lim_{x \rightarrow 0^-} \dfrac{1}{x}=-\infty$: as $x$ gets closer and closer to 0 from the left, $\dfrac{1}{x}$ becomes a larger and larger negative number.

$\displaystyle\lim_{x \rightarrow 3} \dfrac{1}{10}=\dfrac{1}{10}$: no matter what $x$ is, $\dfrac{1}{10}$ is always $\dfrac{1}{10}$. In particular, as $x$ approaches 3, $\dfrac{1}{10}$ stays put at $\dfrac{1}{10}$.

When $x$ is very close to 3, $f(x)$ looks like the function $x^2$. So: $\displaystyle\lim_{x \rightarrow 3} f(x) = \displaystyle\lim_{x \rightarrow 3} x^2=9$

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

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 $$\lim_{x\to +\infty} \frac{4x^3+x}{7x^3 +x^2-2}=\frac{4}{7}.$$