무한대에서의 극한과 점근선

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.

$$\begin{aligned} \lim_{x \to 3}\frac{2x+8}{\frac{1}{x-3}+\frac{1}{x^2-9}}&=\lim_{x \to 3}\frac{2x+8}{\frac{x+3}{x^2-9}+\frac{1}{x^2-9}}\\ &=\lim_{x \to 3}\frac{2x+8}{\frac{x+4}{x^2-9}}\\ &=\lim_{x \to 3}\frac{(2x+8)(x^2-9)}{x+4}=0 \end{aligned}$$