91Ó°ÊÓ

First, we will use the property, \(a^{\log b} = b^{\log a}\), for the given expression in order to simplify it. Let \(y = \left(\frac{1}{3} \cdot 3^{x}+\frac{2}{3} \cdot 2^{x}\right)^{1 / x}\) so that we have: $$y = e^{\ln y} = e^{\left(\frac{1}{x} \ln \left(\frac{1}{3} \cdot 3^{x}+\frac{2}{3} \cdot 2^{x}\right)\right)}$$ Now let's find the limit of the exponent as x approaches \(0\) from the positive: $$\lim _{x \rightarrow 0^{+}}\frac{1}{x} \ln \left(\frac{1}{3} \cdot 3^{x}+\frac{2}{3} \cdot 2^{x}\right)$$ At this stage, we still have an indeterminate form of the type \(\frac{0}{0}\). In order to resolve the indeterminate form, we can use L'Hospital's rule.

We will apply L'Hospital's rule by taking the derivative of both the numerator and denominator of our expression with respect to \(x\). Doing so, we get: $$\lim _{x \rightarrow 0^{+}}\frac{\frac{d}{dx} \ln \left(\frac{1}{3} \cdot 3^{x}+\frac{2}{3} \cdot 2^{x}\right)}{\frac{d}{dx} x}$$ Now compute the derivatives: $$\lim _{x \rightarrow 0^{+}}\frac{\frac{1}{\frac{1}{3} \cdot 3^x + \frac{2}{3} \cdot 2^x} \left(\frac{1}{3} \cdot 3^x \cdot \ln3 + \frac{2}{3} \cdot 2^x \cdot \ln2\right)}{1}$$ Since the limit of the denominator is no longer \(0\), we can substitute \(x = 0\) into the expression directly: $$\frac{\frac{1}{\frac{1}{3} \cdot 3^0 + \frac{2}{3} \cdot 2^0} \left(\frac{1}{3} \cdot 3^0 \cdot \ln3 + \frac{2}{3} \cdot 2^0 \cdot \ln2\right)}{1} = \frac{\frac{1}{1}(\frac{1}{3} \cdot \ln3 + \frac{2}{3} \cdot \ln2)}{1} = \frac{1}{3} \ln3 + \frac{2}{3} \ln2$$

Recall that our original expression was: $$y = e^{\left(\frac{1}{x} \ln \left(\frac{1}{3} \cdot 3^{x}+\frac{2}{3} \cdot 2^{x}\right)\right)}$$ We found the limit of the exponent to be \(\frac{1}{3} \ln3 + \frac{2}{3} \ln2\), so we can substitute it directly: $$y = e^{\left(\frac{1}{3} \ln 3 + \frac{2}{3} \ln 2\right)}$$ Therefore, the value of the limit is: $$\lim _{x \rightarrow 0^{+}}\left(\frac{1}{3} \cdot 3^{x}+\frac{2}{3} \cdot 2^{x}\right)^{1 / x} = e^{\left(\frac{1}{3} \ln 3 + \frac{2}{3} \ln 2\right)}$$