อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Pakpoom
กำหนด$a_n=\sqrt{9n^2+1}+\sqrt{n^2+6n+1}-4n$ ให้หา $\lim_{n \to \infty} a_n$
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$$a_n = \sqrt{9n^2+1}-3n+\sqrt{n^2+6n+1}-n$$$$= \dfrac{1}{\sqrt{9n^2+1}+3n}+\dfrac{6n+1}{\sqrt{n^2+6n+1}+n}$$$$= \dfrac{\frac{1}{n}}{\sqrt{9+\frac{1}{n^2}}+3}+\dfrac{6+\frac{1}{n}}{\sqrt{1+\frac{6}{n}+\frac{1}{n^2}}+1}$$$\displaystyle \lim_{n \to \infty} a_n = \dfrac{0}{6}+\dfrac{6}{2}=3$