Computing $\underset{x\rightarrow0}{\lim}\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}$

By substituting $x=1/n$ and taking the limit from the right we get

$$\lim_{n\to\infty}\left(a^{1/n}+b^{1/n}-c^{1/n}\right)^n$$

Applying Maclaurin series for $a^x=1+x\ln(a)+\mathcal O(x^2)$ we get:

$$a^{1/n}+b^{1/n}-c^{1/n}=1+\frac1n\ln\left(\frac{ab}c\right)+\mathcal O\left(\frac1{n^2}\right)$$

Recall the alternative definition $e^x=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$ and you can show that we hence have:

$$\lim_{n\to\infty}\left(a^{1/n}+b^{1/n}-c^{1/n}\right)^n=e^{\ln(ab/c)}=\frac{ab}c$$

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More formal arguments can be made to show that $e^x=\lim\limits_{n\to-\infty}\left(1+\frac xn\right)^n$ and hence $e^x=\lim\limits_{t\to0}\sqrt[t]{1+xt}$, and furthermore that $e^x=\lim\limits_{t\to0}\sqrt[t]{1+xt+o(t)}$.

$$y=\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}\implies \log(y)=\frac{1}{x}\log\big(a^{x}+b^{x}-c^{x}\big)$$ Now, using $t^x=e^{x \log(t)}$ and using Taylor expansion around $x=0$, we have $$t^x=1+x \log (t)+\frac{1}{2} x^2 \log ^2(t)+O\left(x^3\right)$$ Applying it for $t=a,b,c$,we then have $$a^{x}+b^{x}-c^{x}=1+x (\log (a)+\log (b)-\log (c))+\frac{1}{2} x^2 \left(\log ^2(a)+\log ^2(b)-\log ^2(c)\right)+O\left(x^3\right)$$ Let $$A=(\log (a)+\log (b)-\log (c))\qquad\text{and} \qquad B=\frac{1}{2} \left(\log ^2(a)+\log ^2(b)-\log ^2(c)\right)$$ to make $$a^{x}+b^{x}-c^{x}=1+A x+ B x^2$$ Continuing with Taylor $$\log(1+A x+ B x^2)=A x+ \left(B-\frac{A^2}{2}\right)x^2+O\left(x^3\right)$$ $$\log(y)=A + \left(B-\frac{A^2}{2}\right)x+O\left(x^2\right)$$ which, for sure, shows the limit but also how it is approched. Back to the definition of $A,B$ and simplifying, then, close to $x=0$ $$\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}=\log \left(\frac{a b}{c}\right)+\log \left(\frac{a}{c}\right) \log \left(\frac{c}{b}\right)x+O\left(x^2\right)$$