Limit involving logarithms I have to solve the limit,

$ \displaystyle \lim_{x \to 0} \frac{ \log ((k+x)^{b} - (k-x)^{b} )}{\log x}$

where $k \in (0,1)$ and $b \in (0,1)$ are constant . I have tried using Taylor expansion but it does not work. Thank you.
 A: You could use Newton's Binomial Theorem to say that
$$
\begin{align*}
(k+x)^b-(k-x)^b&=k^b\left[\left(1+\frac{x}{k}\right)^b-\left(1-\frac{x}{k}\right)^b\right]\\
&=k^b\sum_{n=0}^{\infty}\binom{b}{n}\left[\left(\frac{x}{k}\right)^n-\left(-\frac{x}{k}\right)^n\right]\\
&=k^b\sum_{m=0}^{\infty}\binom{b}{2m+1}2\left(\frac{x}{k}\right)^{2m+1}\\
&=2k^b\sum_{m=0}^{\infty}\binom{b}{2m+1}\frac{x^{2m+1}}{k^{2m+1}}\\
&=2k^b\left(\frac{b}{k}x+O(x^3)\right)\\
&=2bk^{b-1}x+O(x^3)\\
&=2bk^{b-1}x(1+O(x^2)),
\end{align*}
$$
where the series statement holds provided $\lvert x\rvert<k$ and the asymptotic bound holds as $x\to0^+$. (And clearly, as $x\to0^+$, we eventually do have $\lvert x\rvert<k$.) But, this says that
$$
\begin{align*}
\log[(k+x)^b-(k-x)^b]&=\log(2k^{b-1}bx)+\log(1+O(x^2))\\
&=\log(2k^{b-1}b)+\log(x)+O(x^2),
\end{align*}
$$
so that
$$
\begin{align*}
\frac{\log[(k+x)^b-(k-x)^b]}{\log x}=\frac{\log(2k^{b-1}b)+\log(x)+O(x^2)}{\log(x)}\to1\text{ as }x\to0^{+}.
\end{align*}
$$
A: it seems that you can use Taylor expansion together with L'Hospital's Rule!
since this limit is $\frac{\infty}{\infty}$ type,so we use the L'Hospital's Rule:
$$lim\frac{\log[(k+x)^b-(k-x)^b]}{\log x}=lim\frac{a^{[(k+x)^b-(k-x)^b]}}{a^x}b((k+x)^{b-1}+(k-x)^{b-1})$$$$=\lim_{x\to 0^+}\left((b(k+x)^{b-1}+b(k-x)^{b-1})\frac{x}{(k+x)^b-(k-x)^b}\right)$$$=lim{\frac{1}{(k+x)^b-(k-x)^b}}$$(xb((k+x)^{b-1}+(k-x)^{b-1})+x^2b(b-1)((k+x)^{b-2}+(k-x)^{b-2})+......$
then,we can use the Taylor expansion as you require:
$f(x)=(xb((k+x)^{b-1}+(k-x)^{b-1})+x^2b(b-1)((k+x)^{b-2}+(k-x)^{b-2})+......$
$lim{\frac{\log[(k+x)^b-(k-x)^b]}{\log x}}=\frac{f(x)}{f(x)}=1$
A: Hint: A L'Hospital's Rule calculation will do it (well, two). To make things simpler, after the first application of L'Hospital's  Rule, separate out the nicely behaved component. 
Added: Use L'Hospital's Rule. So we want
$$\lim_{x\to 0^+}\left((b(k+x)^{b-1}+b(k-x)^{b-1})\frac{x}{(k+x)^b-(k-x)^b}\right).$$
The first term is well behaved as $x\to 0^+$, let's forget about it for a while. Using L'Hospital's Rule on the remaining fraction, we find that its limit is the reciprocal of the limit of the first term! Thus the required limit is $1$.
