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Given the limit:

$$\lim_ {n\to \infty}\left(\frac{n^2+5}{3n^2+1}\right)^{\! n}$$

Is it possible to assume that

$$\lim_ {n\to \infty}\left(\frac{n^2+5}{3n^2+1}\right)^{\! n} = L$$

and then take the natural log of both sides

$$\lim_ {n\to ∞}\left(n \cdot \ln\left(\frac{n^2+5}{3n^2+1}\right)\right) = \ln(L)$$

and keep solving? (I got that $L=0$.)

Fly by Night
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iAmWanteD
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5 Answers5

1

Notice that $L=0$ and $\log 0$ is undefined. However, we can write

$$\lim_{n\to \infty}\left(\frac{n^2+1}{3n^2+1}\right)^n=\lim_{n\to \infty}e^{\log \left(\frac{n^2+1}{3n^2+1}\right)^n}=\lim_{n\to \infty}e^{n\log \left(\frac{n^2+1}{3n^2+1}\right)}=e^{\lim_{n\to \infty}n\log \left(\frac{n^2+1}{3n^2+1}\right)}$$

Noting that the limit in the exponent is $-\infty$, we find

$$\lim_{n\to \infty}\left(\frac{n^2+1}{3n^2+1}\right)^n=0$$

Mark Viola
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1

It's maybe easiest to note that

$$0\le{n^2+5\over 3n^2+1}\le{1\over2}\quad\text{if}\quad n\ge3$$

and therefore, by the Squeeze Theorem,

$$0\le\lim_{n\to\infty}\left(n^2+5\over 3n^2+1\right)^n\le\lim_{n\to\infty}\left(1\over2\right)^n=0$$

Alex M.
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Barry Cipra
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  • Why is the expression always smaller (or equal) to 0.5? – iAmWanteD Sep 18 '15 at 16:13
  • @Osh24: Cross-multiplication gives $2n^2 + 10 \le 3n^2 + 1$, or equivalently $n^2 - 9 \ge 0$ which is true for $n \ge 3$. – Alex M. Sep 18 '15 at 16:18
  • I understand,but why thinking of 1/2 in first place. Why not 1/3, for example? (I understand why the inequality is correct) – iAmWanteD Sep 18 '15 at 16:21
  • @user170231 by the way, maximum is at (0,5) :S – iAmWanteD Sep 18 '15 at 16:24
  • @Osh24 oops I misread the plot as 0.5, not 5.0. Thanks – user170231 Sep 18 '15 at 16:28
  • @AlexM., thank you for the correction. I forgot to multiply the $5$ by $2$ when I cross-multiplied. – Barry Cipra Sep 18 '15 at 16:32
  • @Osh24: Indeed, $\frac 1 3$ is the first thing that you think about; unfortunately, the equality would not be true (check it!), so you try something else. Clearly, it doesn't make sense to try an even smaller number, therefore you're going to try a larger one. Now, what is the first number that is larger than $\frac 1 3$ but still less than $1$? The psychologically natural choice is $\frac 1 2$. – Alex M. Sep 18 '15 at 16:37
  • @Osh24, any number bigger than $1/3$ is eventually bigger than $(n^2+5)/(3n^2+1)$, so you can make the proof work with anything between $1/3$ and $1$. The number $1/2$ is simply convenient. I.e., what Alex M. just said! – Barry Cipra Sep 18 '15 at 16:38
1

First, consider the function $$\mathrm{f}(x) = \frac{x^2+5}{3x^2+1}$$ The derivative is given by using the quotient rule:

\begin{eqnarray*} \mathrm{f}'(x) &=& \frac{2x(3x^2+1)-(x^2+5)(6x)}{(3x^2+1)^2} \\ \\ &=& \frac{-28}{(3x+1)^2} \end{eqnarray*}

This tells us that $\mathrm{f}'(x) < 0$ for all $x \ge 0$, i.e. $\mathrm{f}$ is a decreasing function. This means that $$\mathrm{f}(x) < \mathrm{f}(y) \iff x > y$$ This means that the sequence $\mathrm{f}(n)$, where $n$ is a positive whole number, is strictly decreasing: $$\mathrm{f}(1) > \mathrm{f}(2) > \mathrm{f}(3) > \ldots > \mathrm{f}(n) > \mathrm{f}(n+1) > \ldots$$

Finally, note that when $n =2$ we have $\mathrm{f}(n) = \frac{9}{13}$, and hence $\mathrm{f}(n) < 1$ for all $n \ge 2$.

Coming back to your question. For all $n \ge 2$, we have $$0 < \frac{n^2+5}{3n^2+1} < 1 \implies \lim_{n \to \infty}\left( \frac{n^2+5}{3n^2+1} \right)^{\! n} = 0$$

Fly by Night
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0

Did you try solving the limit directly? I calculated the limit as $0$ as well, but through examining the dominant terms of the numerator and denominator ($n^2$ and $3n^2$), the limit of the inside of the parentheses goes to $\frac{1}{3}$, then raised to the $n^{th}$ power, goes to $0$ as $n$ goes to $\infty$

costrom
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First, yes, your approach is correct.

Second, you could have done it more easily: if $\lim a_n = a$ and $\lim b_n = b$ (with $a,b$ not necessarily finite) and the expression $a^b$ makes sense (in the extended arithmetic that has rules for dealing with infinity), then $\lim \ a_n ^{b_n}$ exists (possibly infinite) and is equal to $a^b$. In your case, this becomes $(\frac 1 3) ^\infty = 0$.

Alex M.
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  • Can I assume that L exists in first place? Because if lim f(x) = L, and L does not exist, it is wrong, isn't it? – iAmWanteD Sep 18 '15 at 15:49
  • @Osh24: No. The existence of $L$ is guaranteed as a consequence of the conditions given in my answer, you don't need to assume it. – Alex M. Sep 18 '15 at 15:57
  • Actually, we cannot take the log of both sides inasmuch as $\log 0$ is not defined. – Mark Viola Sep 18 '15 at 16:10
  • @Dr.MV: Agreed, but please note that I use the term "extended arithmetic", i.e. the thing that is taught in the beginning of a calculus course with $\infty ^\infty = \infty, \frac 1 \infty = 0, \infty + \infty = \infty, \log 0 = -\infty, \log \infty = \infty$ etc. I expect this arithmetic to be basic knowledge to anyone studying limits. – Alex M. Sep 18 '15 at 16:15
  • That's fine. And the approch does lead to the correct answer. ;-)) – Mark Viola Sep 18 '15 at 16:23
  • "Can I assume that L exists in first place? " Many, many students do in problems like this, but it yields an incomplete proof, because logically you've only shown that if the limit exists, then the limit must be ___ . You haven't shown the limit exists though. – zhw. Sep 18 '15 at 16:51
  • @zhw.: I disagree. I show that the limit exists by explicitly constructing (i.e. computing) it. For me, this is an entirely satisfactory argument (from the point of view of its mathematical rigour). – Alex M. Sep 18 '15 at 16:53
  • @zhw: I have edited my answer to clarify your point. – Alex M. Sep 18 '15 at 16:55