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Câu hỏi: Cho khẳng định đúng.
A. $\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln \left( 1+x \right)}{x}=1$.
B. $\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln x}{x}=1$.
C. $\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln \left( 1-x \right)}{x}=1$.
D. $\underset{x\to 0}{\mathop{\lim }} \ln x=1$.
A. $\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln \left( 1+x \right)}{x}=1$.
B. $\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln x}{x}=1$.
C. $\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln \left( 1-x \right)}{x}=1$.
D. $\underset{x\to 0}{\mathop{\lim }} \ln x=1$.
Hướng 1. Ta có $t=\dfrac{1}{x}$. Khi đó
$\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln \left( 1+x \right)}{x}=\underset{x\to 0}{\mathop{\lim }} \left[ \dfrac{1}{x}.\ln \left( 1+x \right) \right]=\underset{x\to \infty }{\mathop{\lim }} \left[ t.\ln \left( 1+\dfrac{1}{t} \right) \right]=\underset{x\to \infty }{\mathop{\lim }} {{\left[ \ln \left( 1+\dfrac{1}{t} \right) \right]}^{t}}=\ln \left[ \underset{x\to \infty }{\mathop{\lim }} {{\left( 1+\dfrac{1}{t} \right)}^{t}} \right]=\ln e=1$.
Hướng 2.
Ta có $\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln \left( 1+x \right)}{x}=\underset{x\to 0}{\mathop{\lim }} \dfrac{{{\left[ \ln \left( 1+x \right) \right]}^{\prime }}}{{{\left( x \right)}^{\prime }}}=\underset{x\to 0}{\mathop{\lim }} \dfrac{1}{1+x}=1$.
$\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln \left( 1+x \right)}{x}=\underset{x\to 0}{\mathop{\lim }} \left[ \dfrac{1}{x}.\ln \left( 1+x \right) \right]=\underset{x\to \infty }{\mathop{\lim }} \left[ t.\ln \left( 1+\dfrac{1}{t} \right) \right]=\underset{x\to \infty }{\mathop{\lim }} {{\left[ \ln \left( 1+\dfrac{1}{t} \right) \right]}^{t}}=\ln \left[ \underset{x\to \infty }{\mathop{\lim }} {{\left( 1+\dfrac{1}{t} \right)}^{t}} \right]=\ln e=1$.
Hướng 2.
Ta có $\underset{x\to 0}{\mathop{\lim }} \dfrac{\ln \left( 1+x \right)}{x}=\underset{x\to 0}{\mathop{\lim }} \dfrac{{{\left[ \ln \left( 1+x \right) \right]}^{\prime }}}{{{\left( x \right)}^{\prime }}}=\underset{x\to 0}{\mathop{\lim }} \dfrac{1}{1+x}=1$.
Đáp án A.