Question

7．$\underset{lim}{x→4}$$\frac{{x}^{3}+x}{{x}^{4}+3{x}^{2}+1}$=$\frac{68}{305}$；$\underset{lim}{x→0}$$\frac{3-\sqrt{9-{x}^{2}}}{{x}^{2}}$=$\frac{1}{6}$．

Answer 1

分析 $\underset{lim}{x→4}$$\frac{{x}^{3}+x}{{x}^{4}+3{x}^{2}+1}$=$\frac{{4}^{3}+4}{{4}^{4}+3×{4}^{2}+1}$=$\frac{68}{305}$，利用洛必達(dá)法則得$\underset{lim}{x→0}$$\frac{3-\sqrt{9-{x}^{2}}}{{x}^{2}}$=$\underset{lim}{x→0}$$\frac{\frac{x}{\sqrt{9-{x}^{2}}}}{2x}$=$\underset{lim}{x→0}$$\frac{1}{2\sqrt{9-{x}^{2}}}$=$\frac{1}{6}$．

解答 解$\underset{lim}{x→4}$$\frac{{x}^{3}+x}{{x}^{4}+3{x}^{2}+1}$=$\frac{{4}^{3}+4}{{4}^{4}+3×{4}^{2}+1}$=$\frac{68}{305}$；
$\underset{lim}{x→0}$$\frac{3-\sqrt{9-{x}^{2}}}{{x}^{2}}$=$\underset{lim}{x→0}$$\frac{\frac{x}{\sqrt{9-{x}^{2}}}}{2x}$
=$\underset{lim}{x→0}$$\frac{1}{2\sqrt{9-{x}^{2}}}$=$\frac{1}{6}$．
故答案為：$\frac{68}{305}$，$\frac{1}{6}$．

點(diǎn)評(píng) 本題考查了極限的定義的應(yīng)用及洛必達(dá)法則的應(yīng)用．