已知椭圆 $C:\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$（$a>0,b>0$）的左右焦点 $F_1$，$F_2$，点 $A,B$ 在椭圆上，满足 $\overrightarrow{AF_2}\cdot\overrightarrow{F_1F_2}=0$，$\overrightarrow{AF_1}=\lambda\overrightarrow{F_1B}$，若椭圆离心率 $e\in\left[ \dfrac{\sqrt{3}}{3},\dfrac{\sqrt{2}}{2} \right]$，则实数 $\lambda$ 的取值范围
解：设点 $A(x_1,y_1)$，$B(x_2,y_2)$
$\because \overrightarrow{AF_2}\cdot\overrightarrow{F_1F_2}=0$
$\therefore x_1=c$
$\therefore \dfrac{c^{2}}{a^{2}}+\dfrac{y_1^{2}}{b^{2}}=1$
$\therefore y_1^{2}=\dfrac{b^{4}}{a^{2}}$
$\because \overrightarrow{AF_1}=\lambda\overrightarrow{F_1B}$
$\therefore (-2c,-y_1)=\lambda(x_2+c,y_2)$
$\therefore \left\{ \begin{array}{l} x_2=-\dfrac{(2+\lambda)c}{\lambda} \\ y_2=-\dfrac{y_1}{\lambda} \end{array} \right.$
$\because \dfrac{x_2^{2}}{a^{2}}+\dfrac{y_2^{2}}{b^{2}}=1$
$\therefore \dfrac{(2+\lambda)^{2}c^{2}}{\lambda^{2}a^{2}}+\dfrac{b^{2}}{\lambda^{2}a^{2}}=1$
$\therefore \dfrac{(2+\lambda)^{2}}{\lambda^{2}}e^{2}+\dfrac{1}{\lambda^{2}}(1-e^{2})=1$
$\therefore e^{2}=\dfrac{\lambda-1}{\lambda+3}\in\left[ \dfrac{1}{3},\dfrac{1}{2} \right]$
$\therefore 3 \leq \lambda \leq 5$