무한대에서의 극한과 점근선

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

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

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.