$\lim\limits_{x\to0}\dfrac{\mathrm{e}^x-1}{3x}$ bằng
| $0$ | |
| $1$ | |
| $3$ | |
| $\dfrac{1}{3}$ |
Chọn phương án D.
Theo quy tắc l'Hospital ta có $$\lim\limits_{x\to0}\dfrac{\mathrm{e}^x-1}{3x}=\lim\limits_{x\to0}\dfrac{\big(\mathrm{e}^x-1\big)'}{(3x)'}=\lim\limits_{x\to0}\dfrac{\mathrm{e}^x}{3}=\dfrac{\mathrm{e}^0}{3}=\dfrac{1}{3}.$$