The limit of $x\frac{a^x\log{a}+b^x\log{b}}{a^x+b^x}-\log{(a^x+b^x)}$ 
Show that
  $$\lim_{x\to\infty}x\frac{a^x\log{a}+b^x\log{b}}{a^x+b^x}-\log{(a^x+b^x)}=0$$ where $a,b>0$.

I have tried L'Hôpital's rule, but makes the function more complicated.
Since $a^x+b^x$ grows faster than $a^x$ and $b^x$, maybe there is some inequality fitting this problem?
What if
$$\lim_{x\to\infty}x\frac{\sum_{i=1}^na_i^x\log{a_i}}{\sum_{i=1}^na_i^x}-\log{(\sum_{i=1}^na_i^x)}$$
 A: If $a=b$, then $L=x\,\frac{a^x\log(a)+b^x\log(b)}{a^x+b^x}-\log(a^x+b^x)=-\log(2)$ and therefore, we have $\lim_{x\to \infty}(-\log(2))=-\log(2)$.  
We now examine the case for which $a\ne b$.  Without Loss of Generality, we may assume that $a>b$.  We can write the term of interest as
$$\begin{align}
L&=x\,\frac{a^x\log(a)+b^x\log(b)}{a^x+b^x}-\log(a^x+b^x)\\\\
&=\frac{a^x\log(a^x)+b^x\log(b^x)-(a^x+b^x)\log(a^x+b^x)}{a^x+b^x}\\\\
&=\frac{(b/a)^x\log((b/a)^x)-\log(1+(b/a)^x)-(b/a)^x\log(1+(b/a)^x)}{1+(b/a)^x} \tag 1
\end{align}$$
Since all of the terms in the numerator approach zero as $x\to \infty$ while the denominator approaches $1$, we have
$$\lim_{x\to \infty}L=0$$
for $a\ne b$.
A: First step: Try a special case.
Setting $a = b$ gives that the entire expression breaks down to $x\log a  - \log(2a^x) = x\log a - \log(2) - x\log a = -\log(2) \neq 0$,
so in general the limit is not zero.
However, if $a < b$ then we see that we can try to get some estimates for how things will grow since, heuristically, $b^x$ trumps $a^x$ in growth.
We see that $$\log(a^x + b^x) - \log(b^x) = \log((a/b)^x + 1)\to \log(1) = 0,$$ since $0 < a/b < 1$. So, we can replace the above difference by 
$$x\left(\frac{a^x\log a + b^x \log b}{a^x + b^x}\right) - x\log(b).$$
This leads to the following computation by putting the $x's$ together and continuing to consolidate and factoring:
$$x\left(\frac{a^x\log a + b^x \log b}{a^x + b^x}\right) - x\log(b) = x\left(\frac{(a/b)^x\log a + \log b}{(a/b)^x + 1} - \log(b)\right) = x\left(\frac{(a/b)^x\log a - (a/b)^x\log b}{(a/b)^x + 1}\right) = x(a/b)^x\left(\frac{\log a - \log b}{(a/b)^x + 1}\right).$$
Now, we see that since $0 < a/b < 1$ that $a(a/b)^x \to 0$ but the other factor converges to $\log(a) - \log(b)$, which is finite. So the entire limit is zero.
Note the continual use of $a$ being strictly less than $b$ in this entire case.
