Đáp án:
$\lim\limits_{x\to 0}\left(e^{8x} + 4x\right)^{\tfrac 1x} = e^{12}$
Giải thích các bước giải:
\(\begin{array}{l}
\quad \lim\limits_{x\to 0}\left(e^{8x} + 4x\right)^{\tfrac 1x}\\
= \lim\limits_{x\to 0}e^{\displaystyle{\ln\left(e^{8x} + 4x\right)^{\tfrac 1x}}}\\
= e^{\displaystyle{\lim\limits_{x\to 0}\ln\left(e^{8x} + 4x\right)^{\tfrac 1x}}}\\
= e^{\displaystyle{\lim\limits_{x\to 0}\dfrac{\ln\left(e^{8x} + 4x\right)}{x}}}\\
= e^{\displaystyle{\lim\limits_{x\to 0}\dfrac{8e^{8x} + 4}{e^{8x} + 4x}}}\quad \text{(Quy tắc L'Hôpital)}\\
= e^{\displaystyle{\dfrac{8e^0 + 4}{e^0 + 4.0}}}\\
= e^{12}
\end{array}\)
