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Why doesn't $\lim\limits_ {n\to \infty}\ (\frac{n+3}{n+4})^n$ equal $1$?

So this is the question.

I found it actually it equals $e^{-1}$. I could prove it, using some reordering and canceling.

However another way I took was this:

$$\lim_ {n\to \infty}\ \left(\frac{n}{n+4}+\frac{3}{n+4}\right)^n$$

with the limit of the first term going to $1$ and the second to $0$. So $(1+0)^n=1$ not $e^{-1}$.

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3  
You set $\frac{n}{n+4}$ to its limiting value $1$ but not the exponent, that is, you are assigning different meanings to the $n$'s inside the parentheses and the $n$ outside the exponent. All the $n$'s must have the same value, no? –  Dilip Sarwate Nov 18 '11 at 20:59
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Do you believe $(A+B)^n$ and $A^n + B^n$ are the same? –  Will Jagy Nov 18 '11 at 21:04
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Then entire expression, both base and exponent, depend on $n$. Why treat the base first, and the exponent later? Or why not try the exponent first, and the base later? Then you would have that for a fixed $n$, $\frac{n+3}{n+4}$ is smaller than $1$, so raised to higher and higher powers will go to $0$. The point is: you can't just decide that you are going to treat the exponent as fixed and deal with the base separately, nor can you decide that you are going to treat the base as fixed and deal with the exponent first; both matter, so they cannot be dealt with separately. –  Arturo Magidin Nov 18 '11 at 21:16
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You might find this helpful: Why is $1^\infty$ considered to be an indeterminate form? –  Mike Spivey Nov 19 '11 at 4:25
    
You cannot mix limits like that $$\lim_{n\to \infty}(\frac{n}{n+4}+\frac{3}{n+4})^n\ne \lim_{n\to \infty}(1+0)^n=1$$ –  AD. Jun 4 at 10:07

4 Answers 4

up vote 31 down vote accepted

Because $1^\infty$ is a tricky beast. Perhaps the power overwhelms the quantity that's just bigger than $1$, but approaching $1$, and the entire expression is large. Or perhaps not...

Perhaps the power overwhelms the quantity that's just smaller than $1$, but approaching $1$, and the entire expression tends to $0$ . Or perhaps not...

In your case, $$ {n+3\over n+4} = 1-{1\over n+4}. $$ And, as one can show (as you did): $$\lim\limits_{n\rightarrow\infty}(1-\textstyle{1\over n+4})^n = \lim\limits_{n\rightarrow\infty}\Bigl[ (1-\textstyle{1\over n+4})^{n+4}\cdot (1-{1\over n+4})^{-4}\Bigr] = e^{-1}\cdot1=e^{-1}.$$

Here, the convergence of $1-{1\over n+4}$ to 1 is too fast for the $n^{\rm th}$ power to drive it back down to $0$.

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I do agree : $1^\infty$ is tricky. You can rewrite it as $\exp(\infty \times \ln(1)) = \exp(\infty \times 0)$ to make the trick even more visible. –  Paul Pichaureau Nov 18 '11 at 21:20
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@David Mindlessly upvoted for the starting sentence. –  Pedro Tamaroff Feb 24 '12 at 6:45

As David Mitra makes clear in his answer, there is a more fundamental question underlying your question, which is:

  • why is $\displaystyle \lim_{n\to \infty} (1 + 1/n)^n$ not equal to $1$?

The same (fallacious) reasoning as the one you give applies. However, this limit is known to equal $e$, not $1$ (and your limit of $e^{-1}$ can easily be obtained from that limit, as David essentially shows).

As David says, there is a tension between the term in parentheses which is tending to $1$ from above, and the power of $n$, which, when applied to any fixed number $>1$, will give larger and larger answers as $n$ increases.

You might want to consider some other related limits to see how they behave, e.g.:

  • $\displaystyle\lim_{n\to\infty} (1+1/n^2)^n$ .

  • $\displaystyle \lim_{n\to\infty} (1 + 1/\log n)^n$ .

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Another way to see:

$\left(\frac{n+3}{n+4}\right)^n=\left(\frac{1+\frac{3}{n}}{1+\frac{4}{n}}\right)^n=\frac{\left(1+\frac{3}{n}\right)^n}{\left(1+\frac{4}{n}\right)^n}.$

Since, $\lim_{n\to\infty}\left(1+\frac{c}{n}\right)^n=e^c$, follows

$$\lim_{n\to\infty}\left(\frac{n+3}{n+4}\right)^n=\lim_{n\to\infty}\frac{\left(1+\frac{3}{n}\right)^n}{\left(1+\frac{4}{n}\right)^n}=\frac{e^3}{e^4}=e^{-1}. $$

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+1 very nice, because you bring it back to the basic defintion. –  draks ... Feb 24 '12 at 7:20

While intuitively it seems "obvious" that $1^\infty$ has to be one, let me point to you that $ \dfrac{n+3}{n+4} <1$. Then the meaning of

$$\left(\frac{n+3}{n+4}\right)^n $$

is a number strictly less than $1$ to a huge power. Now, a number between $0$ and $1$ raised to a larger power gets smaller and smaller, so is it still obvious that it approaches 1?

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