回復 23# mathca 的帖子
第 4 題
設\(P(x,y)\)為雙曲線\(9x^2-16y^2=144\)上一點,且點\(P\)為第一象限內,則\( \displaystyle \lim_{x \to \infty}\sqrt{x |\; 3x-y|\;}\)值為何?
\(\begin{align}
& \underset{x\to \infty }{\mathop{\lim }}\,\sqrt{x\left| 3x-y \right|} \\
& =\underset{x\to \infty }{\mathop{\lim }}\,\sqrt{x\left| 3x-\frac{3}{4}\sqrt{{{x}^{2}}-16} \right|} \\
& =\underset{x\to \infty }{\mathop{\lim }}\,\sqrt{3\left| \frac{4{{x}^{2}}-\sqrt{{{x}^{4}}-16{{x}^{2}}}}{4} \right|} \\
& =\underset{x\to \infty }{\mathop{\lim }}\,\sqrt{3\left| \frac{15{{x}^{4}}+16{{x}^{2}}}{4\left( 4{{x}^{2}}+\sqrt{{{x}^{4}}-16{{x}^{2}}} \right)} \right|} \\
& =\underset{x\to \infty }{\mathop{\lim }}\,\sqrt{3\left| \frac{15{{x}^{2}}+16}{4\left( 4+\sqrt{1-\frac{16}{{{x}^{2}}}} \right)} \right|} \\
& =\infty \\
\end{align}\)