5

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.

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

jimjim
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Paul Manta
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2 Answers2

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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 $1$.

Christian Blatter
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2

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 $$\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n=e?$$

Ross Millikan
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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
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    @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
  • @Paul: Now you are exactly at Ross's answer. Have you seen the e limit before? – davidlowryduda Apr 26 '11 at 18:07
  • @mixedmath and @Ross: Yes, I recognize that limit. But why does my book give $e^4$ as the result? I added the solution given by my book in my question. – Paul Manta Apr 26 '11 at 18:23
  • 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. – Ross Millikan Apr 26 '11 at 18:34