How do we evaluate $\lim_{x\to\infty} \left(\frac{x-c}{x+c}\right)^x$?

Question

$\lim_{x\to\infty} (\frac{x-c}{x+c})^x=\frac{1}{4}$

My teacher did the following steps:

$\lim_{x\to\infty} \ln{(\frac{x-c}{x+c})^x}=\ln{\frac{1}{4}}$

$\lim_{x\to\infty} x\ln{\frac{x-c}{x+c}}=\ln{\frac{1}{4}}$

$\lim_{x\to\infty} x[\ln{(x-c)}-\ln{(x+c)}]=\ln{\frac{1}{4}}$

$\lim_{x\to\infty} x\ln{(x-c)}-x\ln{(x+c)}=\ln{\frac{1}{4}}$

As $x\to\infty, f(x)\to\infty\cdot\infty-\infty\cdot\infty=\infty-\infty$

Then, she did this:

$\lim_{x\to\infty} \frac{\ln{(x-c)}}{\frac{1}{x}}-\frac{\ln{(x+c)}}{\frac{1}{x}}=\ln{\frac{1}{4}}$

At this point, $f(x)\to\frac{\infty}{0}-\frac{\infty}{0}$

But she did L'Hopital's rule here, and when I said you can only do L'Hopital when $f(x)\to\frac{0}{0}$ or $\frac{\infty}{\infty}$, and not when $f(x)\to\frac{\infty}{0}$ she insisted that she was doing the problem correctly. She ended up getting the right answer, but the steps were confusing to me and when I tried to solve it myself, I kept getting stuck.

Can someone please solve the problem step by step and help me out?

Thank you in advance

Answer 1

You should probably fix some value of $c$, as mentioned in the comments. As per the application of the L'Hopital rule:

In your third line, you have $$x[ {\log(x-c)-\log(x+c)}].$$ Rewrite this to $$\frac{\log(x-c)-\log(x+c)}{1/x}.$$ Now, if $x \to \infty$, then both numerator and denominator tend to $0$. So you can apply L'Hopital here, and that should give you the answer. It's a minor thing your teacher missed, or perhaps she skipped over it for pedagogical purposes, so don't think badly of her.

Answer 2

We know that as $x \to \infty$, $\ln(x-c) - \ln(x+c) \to 0$, because it is equal to $\ln\frac{x-c}{x+c}$, which tends to $\ln 1 = 0$. Therefore we can apply L'Hopital's rule to $$\lim_{x \to \infty} \frac{\ln (x-c) - \ln(x+c)}{1/x}$$ even though could not apply it to either $\frac{\ln(x-c)}{1/x}$ or $\frac{\ln(x+c)}{1/x}$ individually. Then we get $$\lim_{x \to \infty} \frac{1/(x-c) - 1/(x+c)}{-1/x^2} = \lim_{x \to \infty} -\frac{2cx^2}{x^2-c^2} = -2c,$$ which we can set equal to $\ln \frac14$ to solve for $c$.

This looks a lot like we got away with applying L'Hopital's rule to the $\frac{\infty}{0}$ terms, but we could only do it because their difference has the form $\frac 00$. (But if you aren't quite following what's going on, or this detail is not explained, that's very confusing!)

Answer 3

Recall that, you can also reach the solution without applying the L'Hôpital's rule :

$$ \small{\begin{align}\lim_{x\to\infty} \left(\frac{x-c}{x+c}\right)^x&=\lim_{x\to\infty} \left(1-\frac {2c}{x+c}\right)^x\\ &=\lim_{x\to\infty} \left(\left(1-\frac{1}{\frac {x+c}{2c}}\right)^{\frac {x+c}{2c}}\right)^{\frac {2cx}{x+c}}\\ &=\lim_{x\to\infty} \left(\left(1-\frac{1}{\frac {x+c}{2c}}\right)^{\frac {x+c}{2c}}\right)^{\frac {2c}{1+\frac cx}}\\ &=\lim_{x\to\infty} \left(\left(1-\frac{1}{\frac {x+c}{2c}}\right)^{\frac {x+c}{2c}}\right)^{\lim_{x\to \infty} {\frac {2c}{1+\frac cx}}}\\ &=e^{-2c}\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align}} $$

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Here, I assume that the question is about :

Find the real number $c$, such that :

$$\lim_{x\to\infty} \left(\frac{x-c}{x+c}\right)^x=\frac 14 $$

holds .

If so, the answer becomes :

$$ \begin{align}e^{-2c}&=\frac 14\\ -2c&=-2\ln 2\\ c&=\ln 2\thinspace \thinspace . \end{align} $$

Answer 4

Here's another way to solve this without L'Hopital. We know that $$\lim_{x\rightarrow\infty}\left(1+\frac1x\right)^x=e$$or$$\lim_{x\rightarrow\infty}\left(1+\frac tx\right)^x=e^t$$ (make the substitution $x\mapsto x/t$). Replace $t$ with $-2c$: $$\lim_{x\rightarrow\infty}\left(1-\frac {2c}x\right)^x=e^{-2c}$$The limit in question can be re-written as $$\lim_{x\rightarrow\infty}\left(1-\frac {2c}{x+c}\right)^x$$and since $x\sim x+c$, the limits are equivalent. So $$e^{-2c}=\frac14\implies c=\ln 2$$