5
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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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    @$P$aul: http://math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form2011-04-26

2 Answers 2

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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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    Fundamentally, the -5 in the denominator doesn't matter when $x$ gets large and the 4 in the numerator becomes the exponent of $e$. The book is getting into the form of the expression for $e$, then moving the limit into the exponential.2011-04-26