Solve $\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}$ I need to solve the limit
$$\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}$$
when $a,b,c \gt 0$.
I'm looking for ways to avoid $\frac{1}{x}$ power.
 A: Note that 
$$
\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}
=\lim_{x \rightarrow 0}\,e^{\left(\displaystyle\frac1x \log(a^x+b^x-c^x)\right)}.
$$
So we only need to calculate 
$$
\lim_{x \rightarrow 0}\frac1x \log(a^x+b^x-c^x).
$$
Since the expression inside the log goes to $1$ as $x\to0$, we can apply L'Hôpital, to get
$$
\lim_{x \rightarrow 0}\frac1x \log(a^x+b^x-c^x)
=\lim_{x\to0}\frac{a^x\,\log a+ b^x\,\log b-c^x\,\log c}{a^x+b^x-c^x}
=\log a+ \log b-\log c.
$$
Then
$$
\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}=e^{\log a+ \log b-\log c}=\frac{ab}c
$$
Added: as suggested by Aryabhata, it remains to justify that one can actually apply L'Hôpital. For the rule to apply, we need to have a quotient $f(x)/g(x)$, with both $f$ and $g$ differentiable at $0$, with $g'(x)\ne0$ on a neighbourhood of $0$, such that $f(0)=g(0)=0$ (less is needed, in fact, only that they go to $0$) and such that $\lim_{x\to0}f'(x)/g'(x)$ exists (it equals $ab/c$ in this case). Here $f(x)=\log(a^x+b^x-c^x)$, $g(x)=x$. The function $x\mapsto a^x+b^x-c^x$ is differentiable everywhere; for $x$ near $0$, the values of this function approximate $1$; on any neighbourhood of $1$ that doesn't contain $0$,  $\log$ is differentiable, and then so is the composition $f(x)=\log(a^x+b^x-c^x)$. And $g'(x)=1\ne0$ for all $x$.
A: Another hint: Let $f(x) = \log (a^x + b^x - c^x)$
What is $f(0)$?
What is $f'(0)$? (The derivative at $0$).
A: HINT:
$$
  \left( a^x + b^x - c^x \right)^\frac{1}{x} = a \left( 1 + \left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x \right)^\frac{1}{x} = a \left( 1 + x \cdot \frac{\left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x}{x} \right)^\frac{1}{x}
$$
Compute $\lim_{x \to 0} \frac{\left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x}{x}$, then use the golden limit $\lim_{x\to 0} \left( 1 + \alpha x\right)^\frac{1}{x}$.
A: Where you start is to realize that $1^\infty$ is an indeterminate form, so you can use L'hopital's rule after you rewrite it as,
$$e^{\left(\lim\limits_{x \rightarrow 0} \frac{\ln(a^x + b^x - c^x)}{x}\right)}$$
A: If you want to avoid L'Hopital: Write $(a^x + b^x - c^x)^{1 \over x}$ as $a(1 + ({b \over a})^x - ({c \over a})^x)^{1 \over x}$, so that it suffices to find
$$\lim_{x \rightarrow 0} (1 +({b \over a})^x - ({c \over a})^x)^{1 \over x}$$
Taking logs, it's enough to find
$$\lim_{x \rightarrow 0} {\ln(1 +({b \over a})^x - ({c \over a})^x) \over x}$$
$$=\lim_{x \rightarrow 0} {\ln(1 +({b \over a})^x - ({c \over a})^x) - \ln(1 +({b \over a})^0 - ({c \over a})^0) \over x - 0}$$
This is the difference quotient for the derivative of $\ln(1 +({b \over a})^x - ({c \over a})^x)$ at $x = 0$. Using the chain rule to find this derivative and plugging in $x = 0$ gives you $\ln(b/a) - \ln(c/a) = \ln(b/c)$. Going backwards, the original limit is ${\displaystyle ae^{\ln(b/c)} = {ab \over c}}$.
