Calculate $\lim_{x\to 0}\left[\left[\log \left(1+\frac{1}{cx}\right)\right]^b-\left[\log \left(1+\frac{1}{x}\right)\right]^b\right]$ I was not able to calculate the limit,
$$\lim_{x\to 0}\left[\left[\log \left(1+\frac{1}{cx}\right)\right]^b-\left[\log \left(1+\frac{1}{x}\right)\right]^b\right]$$
for $c\in ]0,1[$ and $b\in ]0,1]$.
I have tried algebric manipulations, L'Hopital, but nothing works. Any idea?
 A: As $x\to 0$ we have :
\begin{align}
\log\left( 1 +\frac 1 {cx}\right)^b &= \left(-\log (c) - \log (x) + \log(1+cx) \right)^b \\
&= (-\log (x))^b\left( 1 -\frac{\log(1+cx) - \log (c)}{\log(x)}\right)^b \\
&= (-\log(x)^b) \left( 1 + \frac{\log c}{\log x} + o\left(\frac{1}{\log (x)}\right)\right)^b \\
&= (-\log(x))^b \left(1+b\frac{\log c}{\log x} + o\left(\frac{1}{\log(x)}\right)\right)
\end{align}
Replacing $c=1$ we get:
$$\log\left( 1 +\frac 1 {x}\right)^b = (-\log(x))^b \left (1 + o\left(\frac{1}{\log(x)}\right)\right)$$
Therefore :
\begin{align}
\log\left( 1 +\frac 1 {cx}\right)^b - \log\left( 1 +\frac 1 {x}\right)^b &= (-\log(x))^b
 \left(b\frac{\log c}{\log x} + o\left(\frac{1}{\log(x)}\right)\right) \\
&= -b\log (c) (-\log x)^{b-1} + o( (-\log x)^{b-1})
\end{align}
Therefore  :
$$\lim_{x\to 0^+} \left[\log\left( 1 +\frac 1 {cx}\right)^b - \log\left( 1 +\frac 1 {x}\right)^b\right]= \left\{\begin{array}{cc} -\log c &\text{ if } b = 1\\
0 & \text{ if } b<1\end{array}\right.$$
A: Consider the first order expansion
$$\begin{align}
  \log  \left(1+\frac{1}{cx}\right)&=\log \left[\frac{1}{cx}\left(cx+1\right) \right]\\
&= -\log x -\log c +\log(1 + cx)\\
&\approx -\log x -\log c +cx \\
&= -\log x\left(1+\frac{\log c +c x}{\log(x)}\right)\\
\end{align}$$
Also consider $(1+y)^b=1+by +o(y)$
Then
$$\begin{align} \left[\log \left(1+\frac{1}{cx}\right)\right]^b \approx
 (-\log x)^b\left(1+ b\frac{\log c +c x }{\log(x)}\right)
\end{align}
$$
and the difference is
$$ 
(-\log x)^b \frac{b}{\log x}(\log c +(c -1)x))\to (-\log x)^{b-1} b \log c
$$
This tends to zero if $0<b<1$. For $b=1$, it tends to $\log c$.
