ข้อสอบคณิต 9 สามัญ - ปี 2560

$$\begin{gathered} \mathrm{จากโจทย์} a_n = \frac{{\left(3+2n\right)}^{13}{\left(5+n\right)}^2}{{\left(1-2n\right)}^{15}} \\ -\mathrm{หา} \underset{\mathrm{n}\to \infty }{\lim }{\mathrm{a}}_{\mathrm{n}} \\ \mathbf{เลือกพจน์ที่มีดีกรีสูงสุดของแต่ละวงเล็บ} \\ \mathrm{จะได้} {\left(3+2n\right)}^{13} \mathrm{จะได้} {\left(2n\right)}^{13} \\ {\left(5+n\right)}^2 \mathrm{จะได้} n^2 \\ {\left(1-2n\right)}^{15} \mathrm{จะได้} {\left(-2n\right)}^{15} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \underset{n\to \infty }{\lim }{a}_n = \underset{n\to \infty }{\lim }\frac{{\left(2n\right)}^{13}\left(n^2\right)}{{\left(-2n\right)}^{15}} \\ = \underset{n\to \infty }{\lim }\frac{2^{13}{n}^{13}{n}^2}{{\left(-2\right)}^{15}{n}^{{}^{15}}} \\ = \underset{n\to \infty }{\lim }\frac{2^{13}{n}^{15}}{-{\left(2\right)}^{15}{n}^{{}^{15}}} \\ = \frac{-2^{13}}{2^{15}} = \frac{-1}{2^2} \\ = -\frac{1}{4} \end{gathered}$$