12 c

13c

14c

15d

16c

17c

Đáp án:

Câu 10: $A=\lim_{x\to 0}\dfrac{\sqrt[n]{1+ax}-1}{\sqrt[n]{1+bx}-1}$

$=\lim_{x\to 0}\dfrac{(1+ax)^{\dfrac 1n}-1}{(1+bx)^{\dfrac 1m}-1}$

Giới hạn có dạng $\dfrac 00.$ Áp dụng $\text{L'Hospital},$ ta có:

$A=\dfrac{\dfrac 1m(1+ax)^{\dfrac 1n-1}.a}{\dfrac 1m(1+bx)^{\dfrac 1m-1}.b}$

$A=\dfrac{am}{bn}$

$\to C$
Câu 11:

$B=\lim_{x\to 0}\sqrt{1+\alpha x}\sqrt[3]{1+\beta x}\sqrt[4]{1+\gamma x}-1$

$=\lim_{x\to 0}\sqrt{1+\alpha x}\sqrt[3]{1+\beta x}\Big(\sqrt[4]{1+\gamma x}-1\Big)+\sqrt{1+\alpha x}\Big[\big(\sqrt[3]{1+\beta x}-1\big)+\big(\sqrt{1+\alpha x}-1\big)\Big]$

$=\lim_{x\to 0}\Big(\sqrt{1+\alpha x}\sqrt[3]{1+\beta x}\Big)\dfrac{\sqrt[4]{1+\gamma x}-1}x+\lim_{x\to 0}\sqrt{1+\alpha x}\dfrac{\sqrt[3]{1+\beta x}-1}x+\lim_{x\to 0}\dfrac{\sqrt{1+\alpha x}-1}x$

$=\dfrac{\gamma} 4-\dfrac{\beta}3 +\dfrac{\alpha} 2$

Câu 18:

\(N=\lim_{x\to 0}\dfrac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{\sqrt{1+x}-1}\)

\(N=\dfrac{m\sqrt 1-n\sqrt 1}0\)

\(N=-\infty\)