Find the value of the positive constant c such that: $\lim_{x \to \infty}(\frac{x+c}{x-c})^x=10$ I have been trying forever to figure out this problem, but I seem to get stuck in an infinite L'hopitals loop.  See the question below:

Find the value of the positive constant c such that: $\lim_{x \to \infty}(\frac{x+c}{x-c})^x=10$

After rearrainging the problem a few times (mainly because of other indeterminate forms) I get stuck here:
$$\ln10=\lim_{x \to \infty}\frac{\frac{1}{x+c}-\frac{1}{x-c}}{\frac{-1}{x^2}}$$
It seems like there is an infinite loop of L'hopital's rule here, and I have no idea how to get out of it.
Please help!
 A: Compute
$$
\lim_{x\to\infty}x\log\frac{x+c}{x-c}=
\lim_{x\to\infty}\frac{\log(x+c)-\log(x-c)}{1/x}
$$
Now, as you did, this becomes
$$
\lim_{x\to\infty}\frac{\dfrac{1}{x+c}-\dfrac{1}{x-c}}{-\dfrac{1}{x^2}}=
\lim_{x\to\infty}-x^2\left(\dfrac{1}{x+c}-\dfrac{1}{x-c}\right)
$$
which shouldn't be difficult in this form.
A: Directly and without l'Hospital (assuming $\;c\neq 0\;$):
$$\left(\frac{x+c}{x-c}\right)^x=\left(1+\frac{2c}{x-c}\right)^x=\left[\left(1+\frac1{\frac{x-c}{2c}}\right)^{\frac{x-c}{2c}}\right]^{\frac{2cx}{x-c}}\xrightarrow[x\to\infty]{}e^{2c}\ldots$$
A: It is easier without L'Hospital rule. Recall that
$$\lim_{x \to \infty} \left(1+\dfrac{a}x\right)^x = e^a$$
Hence, we have
$$\lim_{x \to \infty} \left(\dfrac{x+c}{x-c}\right)^x = \lim_{x \to \infty} \left(\dfrac{1+c/x}{1-c/x}\right)^x= \dfrac{\lim_{x \to \infty}\left(1+ \dfrac{c}x\right)^x}{\lim_{x \to \infty}\left(1 - \dfrac{c}x\right)^x} = \dfrac{e^c}{e^{-c}} = e^{2c}$$
Hence, we have
$$e^{2c} = 10 \implies c = \dfrac{\ln(10)}2$$
A: $$
\lim_{x\to\infty}\frac{\dfrac{1}{x+c}-\dfrac{1}{x-c}}{-\dfrac{1}{x^2}}=
\lim_{x\to\infty}-x^2\left(\dfrac{1}{x+c}-\dfrac{1}{x-c}\right)
$$
$$= \lim_{x\rightarrow \infty}\frac{2cx^2}{x^2-c^2}$$
dividing both num and den by $x^2$
$$= \lim_{x\rightarrow \infty}\frac{2c}{1-\frac{c^2}{x^2}}$$
$$2c = \ln10$$
$$ c = \frac{\ln10}{2}$$
A: Hint:
$$\lim_{x\to \infty}\left(\cfrac{x+c}{x-c}\right)^x=\lim_{x\to \infty}\left(1+\cfrac{2c}{x-c}\right)^{x-c}\lim_{x\to \infty}\left(1+\cfrac{2c}{x-c}\right)^{c}$$
