# Stuck on Infinite L'hopitals

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.

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$$\lim_{x\to\infty}\frac{\dfrac{1}{x+c}-\dfrac{1}{x-c}}{-\dfrac{1}{x^2}}= \lim_{x\to\infty}-x^2\left(\dfrac{1}{x+c}-\dfrac{1}{x-c}\right)$$

$$= \lim_{x\rightarrow \infty}\frac{2cx^2}{x^2-c^2}$$ dividing both num and den by $x^2$

$$= \lim_{x\rightarrow \infty}\frac{2c}{1-\frac{c^2}{x^2}}$$

$$2c = \ln10$$

$$c = \frac{\ln10}{2}$$

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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.

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So, in that form should I use L'hopital's rule? Because I was always taught that in a 0*infinity situation you were supposed to "move somebody" a.k.a rearrange the equation. –  user97462 Dec 6 '13 at 23:29
@user97462 Transform it into a single fraction; limits at infinity of rational functions are easy, aren't they? –  egreg Dec 6 '13 at 23:31

Directly and without l'Hospital (assuming $\;c\neq 0\;$):

$$\left(\frac{x+c}{x-c}\right)^x=\left(1+\frac{2c}{x-c}\right)^x=\left[\left(1+\frac1{\frac{x-c}{2c}}\right)^{\frac{x-c}{2c}}\right]^{\frac{2cx}{x-c}}\xrightarrow[x\to\infty]{}e^{2c}\ldots$$

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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 $$e^{2c} = 10 \implies c = \dfrac{\ln(10)}2$$

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