Đáp án:
\(\begin{array}{l}
a)\quad \lim\limits_{x\to 0^+}\dfrac{1}{x^{\ln(e^x - 1)}}= 0\\
b)\quad \lim\limits_{x\to 0^+}(x+2^x)^{\tfrac1x}= 2e
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\quad \lim\limits_{x\to 0^+}\dfrac{1}{x^{\ln(e^x - 1)}}\\
= \lim\limits_{x\to 0^+}x^{-\ln(e^x - 1)}\\
= 0^{-\ln(e^0-1)}\\
= 0\\
b)\quad \lim\limits_{x\to 0^+}(x+2^x)^{\tfrac1x}\\
= \lim\limits_{x\to 0^+}e^{\displaystyle{\ln\left[ (x+2^x)^{\tfrac1x}\right]}}\\
= e^{\displaystyle{\lim\limits_{x\to 0^+}\ln\left[ (x+2^x)^{\tfrac1x}\right]}}\\
= e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{\ln(x + 2^x)}{x}}}\\
= e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{\ln2^x + \ln(1 + x.2^{-x})}{x}}}\\
= e^{\displaystyle{\lim\limits_{x\to 0^+}\left[\ln2+\dfrac{\ln(1 + x.2^{-x})}{x}\right]}}\\
= 2e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{\ln(1 + x.2^{-x})}{x}}}\\
= 2e^{\displaystyle{\lim\limits_{x\to 0^+}\dfrac{1 - x\ln2}{2^x + x}}}\qquad (l'Hôpital)\\
= 2e^{\displaystyle{\dfrac{1 - 0.\ln2}{2^0 + 0}}}\\
= 2e
\end{array}\)
