Evaluate $\lim_{x\rightarrow 0^{+}}(3^{x}-2^{x})^{{1}/{x}} $ Evaluate 
$$\lim_{x\rightarrow 0^{+}}(3^{x}-2^{x})^{{1}/{x}}  $$
I tried to use $\ln$ and $e$ with no success.
 A: Hint. As $x$ tends to $0^+$, we have $$\log(3^{x}-2^{x}) \rightarrow -\infty \tag1$$ 
rewriting $$ \begin{align}(3^{x}-2^{x})^{\frac{1}{x}} =e^{\frac 1x \log(3^{x}-2^{x})}\tag2
\end{align}$$ gives
$$ \begin{align}(3^{x}-2^{x})^{\frac{1}{x}}  \sim e^{\frac 1{0^+} \times(-\infty)}\sim e^{(\infty) \times(-\infty)}=0 \tag3
\end{align}$$ and the desired limit is equal to $0$.
A: If $\lim_{x\to 0^+}(3^x-2^x)^{1/x}=L$, then taking $\ln$ of both sides, $\lim_{x\to 0^+}\frac{\ln(3^x-2^x)}{x}=\ln L$. The limit of the numerator is $-\infty$, and of the denominator is $0$, suggesting $\ln L=-\infty$, so that $L=e^{-\infty}=0$. Intuitively this makes sense, since $(3^x-2^x)^{1/x}$, for small $x$, is a small number raised it to a large power, which is close to zero. 
Addendum A better answer: use the squeeze theorem. For $x\le \log_3(3/2)$, we have $3^x<\frac{3}{2}$ and $2^x\ge 1$, so that
$$
0\le (3^x-2^x)^{1/x}\le \left(\frac32-1\right)^{1/x}=\left(\frac12\right)^{1/x}
$$
Then take the limit as $x\to 0$ of the above.
