Đáp án:
\(\begin{array}{l}
1a)\quad \lim\limits_{x\to 0}\dfrac{1-\cos3x}{x^2}= \dfrac92\\
b)\quad\ \ \lim\limits_{x\to 0}\dfrac{\sqrt{1 + 3\sin x} - 1}{\sin2x}= \dfrac34\\
2a)\quad \lim\limits_{x\to 0}\dfrac{\tan(x^2 + x)}{\sin x^3 + 2x}=\dfrac12\\
b)\quad\ \ \lim\limits_{x\to 0}\dfrac{\sqrt{1 + 3\sin x} - 1}{\ln(1+ 3x)}= \dfrac12\\
3)\quad\ \ \lim\limits_{x\to 0}\dfrac{\ln(1 +\tan2x)}{\sin2x + x^2}= 1\\
4)\quad\ \ \lim\limits_{x\to 0}\dfrac{e^{2x} - 1}{\tan x}= 2\\
5a)\quad \lim\limits_{x\to 0}\dfrac{\ln(1 + x^2\sin^2x)}{2x^3\sin x}= \dfrac12\\
b)\quad\ \ \lim\limits_{x\to 0}\dfrac{\ln(2x+5) - \ln5}{4x}= \dfrac{1}{10}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1a)\quad \lim\limits_{x\to 0}\dfrac{1-\cos3x}{x^2}\\
= \lim\limits_{x\to 0}\dfrac{\dfrac12(3x)^2}{x^2}\\
= \dfrac92\\
b)\quad\lim\limits_{x\to 0}\dfrac{\sqrt{1 + 3\sin x} - 1}{\sin2x}\\
= \lim\limits_{x\to 0}\dfrac{\dfrac12\cdot 3\sin x}{2x}\\
=\lim\limits_{x\to 0}\dfrac{\dfrac32x}{2x}\\
= \dfrac34\\
2a)\quad \lim\limits_{x\to 0}\dfrac{\tan(x^2 + x)}{\sin x^3 + 2x}\\
=\lim\limits_{x\to 0}\dfrac{x^2 + x}{x^3 + 2x}\\
=\lim\limits_{x\to 0}\dfrac{x}{2x}\\
=\dfrac12\\
b)\quad \lim\limits_{x\to 0}\dfrac{\sqrt{1 + 3\sin x} - 1}{\ln(1+ 3x)}\\
=\lim\limits_{x\to 0}\dfrac{\dfrac12\cdot3\sin x}{3x}\\
=\lim\limits_{x\to 0}\dfrac{\dfrac32x}{3x}\\
= \dfrac12\\
3)\quad \lim\limits_{x\to 0}\dfrac{\ln(1 +\tan2x)}{\sin2x + x^2}\\
= \lim\limits_{x\to 0}\dfrac{\tan2x}{2x + x^2}\\
= \lim\limits_{x\to 0}\dfrac{2x}{2x}\\
= 1\\
4)\quad \lim\limits_{x\to 0}\dfrac{e^{2x} - 1}{\tan x}\\
= \lim\limits_{x\to 0}\dfrac{2x}{x}\\
= 2\\
5a)\quad \lim\limits_{x\to 0}\dfrac{\ln(1 + x^2\sin^2x)}{2x^3\sin x}\\
= \lim\limits_{x\to 0}\dfrac{x^2\sin^2x}{2x^3\sin x}\\
= \lim\limits_{x\to 0}\dfrac{\sin x}{2x}\\
= \lim\limits_{x\to 0}\dfrac{x}{2x}\\
= \dfrac12\\
b)\quad \lim\limits_{x\to 0}\dfrac{\ln(2x+5) - \ln5}{4x}\\
= \lim\limits_{x\to 0}\dfrac{1}{2(2x+5)}\qquad \text{(l'Hôpital)}\\
= \dfrac{1}{2(2.0+5)}\\
= \dfrac{1}{10}
\end{array}\)
