$\lim_{x\to0}\,(a^x+b^x-c^x)^{\frac1x}$ 
Given $a>b>c>0$, calculate$\displaystyle\,\,\lim_{x\to0}\,(a^x+b^x-c^x)^{\frac1x}\,$

I tried doing some algebraic manipulations and squeeze it, but couldn't get much further.
 A: $$ \begin{align} 
\color{blue}{L} &= \lim_{x\to0}\left(a^x+b^x-c^x\right)^{\frac1x} \\[3mm] 
\color{red}{\log{L}} &= \lim_{x\to0}\frac{\log\left(a^x+b^x-c^x\right)}{x}=\frac{0}{0} \qquad\left\{\log\left(a^0+b^0-c^0\right)=\log(1)\right\} \\[3mm] 
&= \lim_{x\to0}\frac{\frac{d}{dx}\left[\,\log\left(a^x+b^x-c^x\right)\,\right]}{\frac{d}{dx}\left[\,x\,\right]} \\[3mm] 
&= \lim_{x\to0}\frac{a^x\log{a}+b^x\log{b}-c^x\log{c}}{a^x+b^x-c^x} \\[3mm] 
&= \log{a}+\log{b}-\log{c}=\color{red}{\log{\frac{a\,b}{c}}} \quad\Rightarrow\, \color{blue}{L=\frac{a\,b}{c}}
\end{align} 
$$
A: There's no need for L'Hopital. Let $f(x) = a^x+b^x-c^x.$ If we apply $\ln$ to our expression, we get
$$\tag 1 \frac{\ln f(x)}{x} = \frac{\ln f(x)-\ln f(0)}{x-0}.$$
By definition of the derivative, $(1) \to (\ln f)'(0)$ as $x\to 0.$ Thus the limit of $(1)$ equals
$$\frac{f'(0)}{f(0)}  = \frac{\ln a + \ln b - \ln c}{1} = \ln (ab/c).$$
Exponentiating back gives $ab/c$ for the original limit.
A: write your term in the form $$e^{\lim_{x \to 0}\frac{\ln(a^x+b^x-c^x)}{x}}$$
and use L'Hospital
distinguish two cases:
$x$ tends to $0^+$ and $x$ tends to $0^-$ 
A: $$\lim_{x\to0}(a^x+b^x-c^x)^\frac{1}{x}=[1^\infty]=\exp\lim_{x\to 0}(a^x+b^x-c^x-1)\frac{1}{x}\boxed=\\(a^x+b^x-c^x-1)\frac{1}{x}=(a^x-c^x+b^x-1)\frac{1}{x}=c^x\cdot\frac{\left(\frac{a}{c}\right)^x-1}{x}+\frac{b^x-1}{x}\\ \boxed =\exp \lim_{x\to 0}\left(c^x\cdot\frac{\left(\frac{a}{c}\right)^x-1}{x}+\frac{b^x-1}{x} \right)=\exp\left(\ln\frac{a}{c}+\ln b\right)=\exp\left( \ln \frac{ab}{c} \right)=e^\frac{ab}{c}$$
