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This question already has an answer here:

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 $\frac{\infty}{\infty}$?

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marked as duplicate by Jyrki Lahtonen, Namaste calculus Dec 7 '18 at 14:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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HINT

Note that the inverse

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

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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 $e \times 1 = e$.

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

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    $\begingroup$ though this does not use $e^{\log_e\left(\left(\frac{x}{x-1}\right)^x \right)} = e^{x\log_e\left(\frac{x}{x-1}\right)}$ $\endgroup$ – Henry Dec 7 '18 at 1:01
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    $\begingroup$ +1 You beat me by a minute while I was writing the same answer. $\endgroup$ – Ethan Bolker Dec 7 '18 at 1:01
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Remember, for continuous function $f$, we can say, when we compose the function with some other function $g$,

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

In your case,

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

Alternatively (or in conjunction with this), in line with what other people have suggested, you could also observe

$$\frac{x}{x-1} = \frac{x-1+1}{x-1} = 1 + \frac{1}{x-1}$$

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