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

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 $e^x=\lim\limits_{t\to0}\sqrt[t]{1+xt+o(t)}$.
A: $$y=\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}\implies \log(y)=\frac{1}{x}\log\big(a^{x}+b^{x}-c^{x}\big)$$
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 $x=0$
$$\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}=\log \left(\frac{a b}{c}\right)+\log \left(\frac{a}{c}\right) \log
   \left(\frac{c}{b}\right)x+O\left(x^2\right)$$
A: More generally,
if
$f(x)
=\left(\sum_{k=1}^n c_ia_i^x\right)^{1/x}
$,
then
$f(x)
=e^{g(x)}
$
where
$g(x)
=\ln\left(\sum_{k=1}^n c_ia_i^x\right)^{1/x}
=\frac1{x}\ln\left(\sum_{k=1}^n c_ia_i^x\right)
$.
Then,
as $x \to 0$,
$\begin{array}\\
g(x)
&=\frac1{x}\ln\left(\sum_{k=1}^n c_ia_i^x\right)\\
&=\frac1{x}\ln\left(\sum_{k=1}^n c_ie^{\ln(a_i)x}\right)\\
&=\frac1{x}\ln\left(\sum_{k=1}^n c_i(1+\ln(a_i)x+O(x^2))\right)\\
&=\frac1{x}\ln\left(\sum_{k=1}^n c_i+x\sum_{k=1}^n c_i\ln(a_i)+O(x^2))\right)\\
&=\frac1{x}\ln (C\left(1+\frac{x}{C}\sum_{k=1}^n c_i\ln(a_i)+O(x^2)\right))
\qquad C=\sum_{k=1}^n c_i\\
&=\frac1{x}(\ln (C)+\ln\left(1+\frac{x}{C}\sum_{k=1}^n c_i\ln(a_i)+O(x^2)\right))\\
&=\frac1{x}(\ln (C)+\frac{x}{C}\sum_{k=1}^n c_i\ln(a_i)+O(x^2))\\
&=\frac{\ln(C)}{x}+\frac{1}{C} \ln(\prod_{k=1}^na_i^{c_i})+O(x)\\
\text{if }C=1
&\text{as in this case }(C=1+1-1=1)\\
g(x)
&= \ln(\prod_{k=1}^na_i^{c_i})+O(x)\\
\text{so}\\
f(x)
&e^{g(x)}\\
&=\prod_{k=1}^na_i^{c_i}+O(x)\\
\text{if } C > 1\\
g(x)
&\to \infty\\
\text{so}\\
f(x)
&\to \infty\\
\text{if } C < 1\\
g(x)
&\to -\infty\\
\text{so}\\
f(x)
&\to 0\\
\end{array}
$
A: 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 $\dfrac{ab}{c}.$
