Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to0^+}\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)^{1/x}=\sqrt[3]{12}.$$
To solve this limit we will transform the expression under the limit. Since the natural exponential function and the natural logarithmic function are inverse to each other we have that $c=e^{\ln c}.$ This gives $\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)^{1/x}=e^{\ln \left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)^{1/x}}=e^{\frac{1}{x}\ln\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)}$ where we used the logarithmic rule $\ln b^c=c\ln b$. Also because the exponential function is continuous it can exchange places with the limit so we get $$L=\lim_{x\to0^+}\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)^{1/x}=\lim_{x\to0^+}e^{\frac{1}{x}\ln\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)}=e^{\lim_{x\to0}{\frac{1}{x}\ln\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)}}=e^l,$$ where we have denoted $l=\lim_{x\to0^+}{\frac{1}{x}\ln\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)}$ Now let's calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$l=\lim_{x\to0^+}{\frac{1}{x}\ln\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)}=\lim_{x\to0^+}\frac{\ln\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)}{x}=\left[\frac{\ln\left(\frac{1}{3}3^{0^+}+\frac{2}{3}2^{0^+}\right)}{0^+}\right]=\left[\frac{\ln\left(\frac{1}{3}+\frac{2}{3}\right)}{0}\right]=\left[\frac{\ln1}{0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to0^+}\frac{\left(\ln\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)\right)'}{(x)'}=\lim_{x\to0^+}\frac{\frac{1}{\frac{1}{3}3^x+\frac{2}{3}2^x}\left(\frac{1}{3}3^x+\frac{2}{3}2^x\right)'}{1}=\lim_{x\to0^+}\frac{\frac{1}{3}3^x\ln3+\frac{2}{3}2^x\ln2}{\frac{1}{3}3^x+\frac{2}{3}2^x}=\left[\frac{\frac{1}{3}3^0\ln3+\frac{2}{3}2^0\ln2}{\frac{1}{3}3^0+\frac{2}{3}2^0}\right]=\frac{1}{3}\ln3+\frac{2}{3}\ln2=\frac{1}{3}\ln12.=\ln\sqrt[3]{12}.$$ Returning this into the original limit we get $$L=e^l=e^{\ln\sqrt[3]{12}}=\sqrt[3]{12}.$$