Compute the limit: $\lim_{x\to \infty}x\ \log\left(\frac{x+c}{x-c}\right)$ Does anyone know how o compute the following limit?
$$\lim_{x\to \infty}x\ \log\left(\frac{x+c}{x-c}\right)$$
I tried to split it up as follows:
$\lim_{x\to \infty}\left[x\ \log(x+c)-x\ \log(x-c)\right]$ but still this is indeterminate form of type $\infty-\infty$. Any thoughts please?
 A: Using L'Hospital:
$$\lim_{x\to \infty}x\ \log\left(\frac{x+c}{x-c}\right)=\lim_{x\to \infty}\frac{\log\left(\frac{x+c}{x-c}\right)}{\frac 1x}\underbrace{=}_{\mathrm{L'Hospital}} \lim_{x\to \infty} \frac{\frac{x-c}{x+c}\cdot \frac{-2c}{(x-c)^2}}{-\frac {1}{x^2}}=\lim_{x\to \infty} \frac{x-c}{x+c}\cdot \frac{2cx^2}{(x-c)^2}=2c$$
Without using L'Hospital:
$$\lim_{x\to \infty}x\ \log\left(\frac{x+c}{x-c}\right)=\lim_{x\to \infty} \log\left(1+\frac{2c}{x-c}\right)^x=\log \lim_{x\to \infty}\left(1+\frac{2c}{x-c}\right)^x\\=\log \lim_{x\to \infty}\left(\left(1+\frac{2c}{x-c}\right)^{x-c}\right)^{\frac{x}{x-c}}=\log e^{2c}=2c.$$
A: We have:$$\lim_{x\to \infty}x\ \log\left(\frac{x+c}{x-c}\right) = \lim_{x\to \infty}\ \log\left(\frac{x+c}{x-c}\right)^x = \log \lim_{x \to +\infty}\left(\frac{1+\frac{c}{x}}{1-\frac{c}{x}}\right)^x = \log \lim_{x \to +\infty} \frac{\left(1+\frac{c}{x}\right)^x}{\left(1-\frac{c}{x}\right)^x}$$
Using the fundamental limit for $e$: $$= \log \frac{e^c}{e^{-c}} =  \log e^{2c} = 2c.$$
A: Let $u = \dfrac{c}{x}$, then: 
Note that: $x\log\left(\dfrac{x+c}{x-c}\right) = c\dfrac{\log\left(\dfrac{1+u}{1-u}\right)}{u} = c\left(\dfrac{\log(1+u)}{u} - \dfrac{\log(1-u)}{u}\right) \to c(1-(-1)) = 2c$ since $x \to \infty \iff u \to 0$.
A: Hint; Let $t=\frac1x$${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$
