Đáp án:a)1
b)$\frac{1}{3}$
c)$-\infty$
d)$-\infty$
Giải thích các bước giải:
a)$\lim_{x\rightarrow 0}\frac{x^{n}-1}{x^{m}-1}=\frac{-1}{-1}=1$
b)$D=\lim_{x\rightarrow 0}\frac{\sqrt[3]{x+1}-1}{\sqrt[4]{2x+1}-1}$
=$\lim_{x\rightarrow 0}\frac{x+1-1(\sqrt[4]{2x+1}+1)}{2x+1-1(\sqrt[3]{x+1}+1)}$
=$\lim_{x\rightarrow 0}\frac{x(\sqrt[4]{2x+1}+1)}{2x(\sqrt[3]{x+1}+1)}$
=$\lim_{x\rightarrow 0}\frac{\sqrt[4]{2x+1}+1}{2(\sqrt[3]{x+1}+1)}$
=$\frac{1}{3}$
c)$B=\lim_{x\rightarrow 1}\frac{\sqrt{5+4x}-\sqrt[3]{7+6x}}{x^{3}+x^{2}-x-1}$
=$\lim_{x\rightarrow 1}\frac{\sqrt{5+4x-7-6x}}{(x^{3}+x^{2}-x-1)(\sqrt{5+4x}+\sqrt[3]{7+6x})}$
=$\lim_{x\rightarrow 1}\frac{-2(x+1)}{(x^{2}-1)(x+1))(\sqrt{5+4x}+\sqrt[3]{7+6x})}$
=$\lim_{x\rightarrow 1}\frac{-2}{(x^{2}-1)(\sqrt{5+4x}+\sqrt[3]{7+x})}$
=$-\infty$
d)$M=\lim_{x\rightarrow 0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^{2}}$
=$\lim_{x\rightarrow 0}\frac{1+4x-1-6x}{x^{2}(\sqrt{1+4x}+\sqrt[3]{1+6x})}$
=$\lim_{x\rightarrow 0}\frac{-2x}{x^{2}(\sqrt{1+4x}+\sqrt[3]{1+6x})}$
=$\lim_{x\rightarrow 0}\frac{-2}{x(\sqrt{1+4x}+\sqrt[3]{1+6x})}=-\infty$
