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