Computing $\underset{x\rightarrow0}{\lim}\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}$ A friend asked me to help him with calculating a certain hideous limit:
$$\underset{x\rightarrow0}{\lim}\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x},\space\space0<a,b,c\in\mathbb{R}$$
I came up with a solution (and wolfram alpha confirmed), but unfortunately it involves a lot of steps and a couple of theorems and identities so we're actually reaching out hoping someone can come up with a better solution. Hopefully a more intuitive one!
Here's what I had in mind:
$$\underset{x\rightarrow0}{\lim}\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}=\underset{x\rightarrow0}{\lim}e^{\ln{\big((a^{x}+b^{x}-c^{x})^\frac{1}{x}\big)}}=\underset{x\rightarrow0}{\lim}e^\frac{\ln{\big(a^{x}+b^{x}-c^{x}\big)}}{x}$$
Since $e^x$ is continuous and therefore we have:
$$\underset{x\rightarrow{x_0}}{\lim}e^{f(x)}=e^{\underset{x\rightarrow{x_0}}{\lim}f(x)}$$
So we focus on finding:
$$\underset{x\rightarrow0}{\lim}\frac{\ln{\big(a^{x}+b^{x}-c^{x}\big)}}{x}$$ 
Suppose it is equal to some $L$, then our original limit will be equal to $e^L$.
We now note that:
$$\underset{x\rightarrow0}{\lim}x=0\space,\space \underset{x\rightarrow0}{\lim}\ln{\big(a^{x}+b^{x}-c^{x}\big)}=\ln{(a^0+b^0-c^0)}=\ln{(1+1-1)}=\ln(1)=0$$
So our limit takes the indeterminate form  $"\frac{0}{0}"$. After checking all the conditions are satisfied we proceed with L'Hospital setting $g(x)=x$ and $f(x)=\ln{\big(a^{x}+b^{x}-c^{x}\big)}$, and get the following: (oh boy this is going to be an ugly)
$$\underset{x\rightarrow0}{\lim}\frac{f(x)}{g(x)}=\underset{x\rightarrow0}{\lim}\frac{f'(x)}{g'(x)}=\underset{x\rightarrow0}{\lim}\frac{\frac{a^{x}\ln{(a)}+b^{x}\ln{(b)}-c^{x}\ln{(c)}}{a^{x}+b^{x}-c^{x}}}{1}=\ln{(a)}+\ln{(b)}-\ln{(c )}=\ln{\big(\frac{ab}{c}\big)}$$
(Again, the computation was straightforward because $\ln$ is continuous and the limit of composition of functions).
So we finally get:
$$\underset{x\rightarrow0}{\lim}\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}=\underset{x\rightarrow0}{\lim}e^\frac{\ln{\big(a^{x}+b^{x}-c^{x}\big)}}{x}=e^{\ln{(\frac{ab}{c})}}=\frac{ab}{c}$$
And that's  it! As you can see it's not that intuitive, and it involves a lot of computation and some theorems and identities and as a result- many steps. 
I would appreciate in other insight regrading formality and other perspectives on calculating this limit. Thank you

Apply $\ln$ to the expression to get
$$\tag 1 \frac{\ln(a^x+b^x-c^x)}{x}.$$
Let $f(x) = \ln(a^x+b^x-c^x).$ Then $(1)$ equals
$$\frac{f(x)-f(0)}{x}.$$
The limit of this as $x\to 0$ is, by definition, $f'(0).$ Let's compute:
$$f'(x) = \frac{1}{a^x+b^x-c^x}\cdot (\ln a\cdot a^x + \ln b\cdot b^x -\ln c\cdot c^x).$$
Thus $f'(0)= \ln a+ \ln b -\ln c = \ln \left(\dfrac{ab}{c}\right).$ Exponentiating back shows the original limit is $\dfrac{ab}{c}.$
