Đáp án:
\(\lim_{x\to 8^+}\dfrac{3\sqrt x+\sqrt{x-8}-2}{\sqrt{x^2-64}}=\dfrac 14\)
Giải thích các bước giải:
Ta có: \(\dfrac{3\sqrt x+\sqrt{x-8}-2}{\sqrt{x^2-64}}\)
\(=\dfrac 1{\sqrt{x+8}}+\dfrac{x-8}{\left[\left(\sqrt[3]x\right)^2+2\sqrt[3]x+4\right]\cdot \sqrt{x^2-64}}\)
\(=\dfrac 1{\sqrt{x+8}}+\dfrac{\sqrt{x-8}}{\left[\left(\sqrt[3]x\right)^2+2\sqrt[3]x+4\right]\cdot \sqrt{x+8}}\)
Do đó: \(\lim_{x\to 8^+}\dfrac{3\sqrt x+\sqrt{x-8}-2}{\sqrt{x^2-64}}=\lim_{x\to 8}\left\{\dfrac 1{\sqrt{x+8}}\dfrac{\sqrt{x-8}}{\left[\left(\sqrt[3]x\right)^2+2\sqrt[3]x+4\right]\cdot \sqrt{x+8}}\right\}\)
\(=0+\dfrac 1{\sqrt{8+8}}\)
\(=\dfrac 1{\sqrt{16}}\)
\(=\dfrac 14\)
