Question

5．求函数极限：$\underset{lim}{x→4}$$\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}$．

Answer 1

分析 由洛必达法则可得$\underset{lim}{x→4}$$\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}$=$\frac{2\sqrt{2}}{3}$．

解答 解：$\underset{lim}{x→4}$$\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}$
=$\underset{lim}{x→4}$$\frac{\frac{2}{2\sqrt{2x+1}}}{\frac{1}{2\sqrt{x-2}}}$
=$\frac{2}{2×3}$×2$\sqrt{2}$=$\frac{2\sqrt{2}}{3}$．

点评 本题考查了函数的极限的求法及应用．