# Evaluating $\lim_{x\to \infty}(\frac{x}{x-1})^x$ [duplicate]

I am going over a solution given to solving the follow limit, $$\lim_{x\to \infty}(\frac{x}{x-1})^x$$ The solution continues as follows,

Consider raising the function to $$e^{ln\cdots}$$

We can find the limit as follows, $$\lim_{x\to \infty} x \ln(\frac{x}{x-1}) = \lim_{x\to \infty} \frac{\ln(\frac{x}{x-1})}{\frac{1}{x}}$$

The solution argues this is just $$\frac{0}{0}$$ and as such we can apply L'Hospital's rule. It continues on to find the limit equals 1, so the limit of the function is $$e$$.

However, I don't understand how that expression evaluates to $$\frac{0}{0}$$, in fact it seems to express $$\frac{\ln(\frac{\infty}{\infty})}{0}$$ I assume the argument is that $$\frac{\infty}{\infty}$$ equals 1, and $$\ln(1) = 0$$, so we have $$\frac{0}{0}$$. But I thought we cannot evaluate $$\frac{\infty}{\infty}$$?

## marked as duplicate by Jyrki Lahtonen, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 7 '18 at 14:12

• $\ln \left(\frac{x-1}{x}\right)=\ln \left(1-\frac{1}{x}\right)$. As $x \to \infty$, the expression approaches $\ln 1=0$. – Anurag A Dec 7 '18 at 0:47
• Do the answerers seriously think that this limit would not have been handled multiple times here already? Shame on you! – Jyrki Lahtonen Dec 7 '18 at 10:27

No, you do not need to have any conclusions about $$\ln(\infty/\infty)$$ here (and those wouldn't work anyways!). To see that this is indeed a $$0/0$$ form, go a bit more carefully through the logarithm: We have

$$\lim_{x \to \infty} \ln \left(\frac{x - 1}{x}\right) = \lim_{x \to \infty} \ln \left(1 - \frac 1 x\right) = \ln 1 = 0$$

which is what you need.

HINT

Note that the inverse

$$\left(\frac{x-1}{x}\right)^x=\left(1-\frac{1}{x}\right)^x$$

L'Hopital is rarely the method of choice. In this case, let $$y = x-1$$. Then $$\left(\frac{x }{x-1}\right)^x = \left(\frac{y+1}{y}\right)^{y+1} = \left(1 + \frac{ 1}{y}\right)^{y }\left(1 + \frac{ 1}{y}\right)^{ 1}.$$ Now you can recognize the limit as $$y \to \infty$$ as $$e \times 1 = e$$.

$$\lim\limits_{x\to \infty}\left(\frac{x}{x-1}\right)^x = \lim\limits_{y\to \infty}\left(\frac{y+1}{y}\right)^{y+1} = \lim\limits_{y\to \infty}\left(1 +\frac{1}{y}\right)\lim\limits_{y\to \infty}\left(1+\frac{1}{y}\right)^{y} = 1 \times e = e$$

• though this does not use $e^{\log_e\left(\left(\frac{x}{x-1}\right)^x \right)} = e^{x\log_e\left(\frac{x}{x-1}\right)}$ – Henry Dec 7 '18 at 1:01
• +1 You beat me by a minute while I was writing the same answer. – Ethan Bolker Dec 7 '18 at 1:01

Remember, for continuous function $$f$$, we can say, when we compose the function with some other function $$g$$,

$$\lim_{x \to c} f(g(x)) = f \left( \lim_{x \to c} g(x) \right)$$

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