# Limit $\lim_{x\to \infty} \left(\frac{f(x+1)}{f(x)}\right)^x$

Can someone please explain to me how to solve this? According to my book the result should be $e^4$, however I cannot understand the proposed solution. Can someone please take the time to walk me through it?

$$f : \mathcal R \mapsto \mathbb R, f(x) = (x - 2)(x - 3)(x - 4)(x - 5)$$ $$\lim_{x\to \infty} \left(\frac{f(x+1)}{f(x)}\right)^x$$

Edit: Partial solution.

I can get up to the following point. From here onwards however I do not know how to continue in order to get $e^4$. It appears to me that the result is $1^\infty = 1$ at this point (but that's not the case according to my book):

$$\lim_{x\to \infty} \left(\frac{x-1}{x-5}\right)^x$$

Edit 2: Solution given by my book.

$$\lim_{x\to \infty} \left(1+\frac{4}{x-5}\right)^x$$ $$= \lim_{x\to \infty} \left(\left(1+\frac{4}{x-5}\right)^\frac{x - 5}{4}\right)^{\frac{4}{x - 5}x}$$ $$= e^{\lim_{x\to \infty} \frac{4x}{x - 5}} = e^4$$

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you probably mean $\lim (f(x+1)/f(x))^x$, if you want to get that answer –  user8268 Apr 26 '11 at 17:46
Yes, sorry for the mistake. :( –  Paul Manta Apr 26 '11 at 17:48
@Paul $1^{\infty}$ is an indeterminate form. You can't directly substitute to find its limit. –  Uticensis Apr 26 '11 at 17:52
@Billare Ah, I suspected so, but Wolfram|Alpha told me otherwise so I assumed $1^\infty$ isn't indeterminate. –  Paul Manta Apr 26 '11 at 17:55
–  JavaMan Apr 26 '11 at 18:01

You are here at an "archetypic" experience you have to face when going into mathematics. It all starts with the search for $$\lim_{n\to\infty}\Bigl( 1+{1\over n} \Bigr)^n$$ (you may write $x$ instead of $n$). If the inner $n$ goes to $\infty$ first, the limit is $1$, and if the $n$ in the exponent goes to $\infty$ first, the limit is $\infty$. As a matter of fact (and this has to be proven the hard way) the true limit is a finite number, namely Euler's number $e\doteq 2.718$. Accepting this, it is easy to show that for any fixed $y>0$ one has $$\lim_{x\to\infty}\Bigl( 1+{y\over x} \Bigr)^x=e^y\ .$$ The $x-5$ in your denominator causes no trouble: Just adapt the exponent accordingly, and the extra factor with constant exponent $5$ will converge to $1$.

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Hint: What do you get when you write the fraction $\frac{f(x+1)}{f(x)}$ just as a function of $x$ by substituting in $x+1$ in the numerator?

Added in response to the edit: $\frac{x-1}{x-5}=1+\frac{4}{x-5}$, so you are looking for $$\lim_{x\to \infty} \left(1+\frac{4}{x-5}\right)^x.$$ Have you seen $$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e?$$

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The problem I'm facing isn't that basic. I'll add some more details so people won't bother trying to explain the very basic things to me. :) –  Paul Manta Apr 26 '11 at 17:49
@Paul: Then you should show the work you have done, such as what you have for the function you are taking the limit of. It will be easier to answer your question. –  Ross Millikan Apr 26 '11 at 17:51
@Ross I edited my question. –  Paul Manta Apr 26 '11 at 17:53
@Paul Hint: Try converting $\frac{x-1}{x-5}$ into a partial fraction. –  Tyler Apr 26 '11 at 17:54
@Tyler You mean something like $1 + \frac{4}{n - 5}$? I still don't know how to continue. –  Paul Manta Apr 26 '11 at 17:57