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I'm trying to find out the limits of a sequence of the type $\frac{\infty}{\infty}$, but I got stuck and am starting to get frustrated. The sequence in question is: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}$$ I tried looking at its parts and figuring out their own limits to try and help me, but I didn't get anywhere useful. I tried rationalizing the square roots and this is what I have at the moment: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}\times \frac{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}} = \lim_{n \to \infty}\frac{(n^2+3n+1)-(n^2+3n-1)}{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}=\lim_{n \to \infty}\frac{2}{{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}}$$

RandomJoe
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    Your question is more of the type $\frac{0}{0}$ and so far you have manged to transform it to $\frac{2}{0\times \infty}$. Note that $\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1} = 2n+o(n)$ for large $n$ – Henry Mar 27 '17 at 16:06

2 Answers2

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Observe that $$ \lim_{n\to\infty}(\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1})= \lim_{n\to\infty}\frac{2}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}=0 $$ as well as $$ \lim_{n\to\infty}(\ln(1+n)-\ln(n+2))= \lim_{n\to\infty}\ln\frac{1+n}{n+2}=0 $$ so your limit is in the form $0/0$, rather than $\infty/\infty$.

On the other hand, you can observe that $$ \lim_{n\to\infty}n(\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1})= \lim_{n\to\infty}\frac{2n}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}=1 $$ and that $$ \lim_{n\to\infty}n(\ln(1+n)-\ln(n+2))= \lim_{n\to\infty}n\ln\frac{1+n}{n+2} $$ is finite as well: with $t=1/n$, the limit becomes $$ \lim_{t\to0^+}\frac{\ln(t+1)-\ln(1+2t)}{t}=-1 $$

So you can write your limit as $$ \lim_{n \to \infty}\frac{n(\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1})}{n(\ln(1+n) - \ln(2+n))} $$ and finish up.

egreg
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  • I understand that I should've stuck with my approach to look at the different components. I'm still not really sure why are we multiplying both sides of the fraction by $n$ ? How did you come to this decision? The other thing that is confusing to me is that using your solution I get $-1$ but the answer I have been given is $-2$. I'm I still doing something wrong or is the answer wrong? – RandomJoe Mar 27 '17 at 18:47
  • @RandomJoe You can see in the first limit that the denominator is essentially like $2n$, so if we mu ltiply by $n$, we get a finite limit. – egreg Mar 27 '17 at 19:39
  • I think you have a mistake, we have $\lim_ {t \to 0} {\log(1+t)-\log(1+2t) \over t} = -1$. – copper.hat Mar 29 '17 at 05:50
  • @copper.hat Yes, you're right – egreg Mar 29 '17 at 06:06
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I get the limit to be $-1$.

Note that $\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)} = \frac{ n^2( \sqrt{1+{3 \over n} + {1 \over n^2}} - \sqrt{1+{3 \over n} - {1 \over n^2}})}{n\ln(\frac{1+{1 \over n}}{1 + {2 \over n}})}$.

We have $\lim_{x \to 0} {1 \over x} \ln(\frac{1+x}{1 + 2x}) = -1 $ and $\lim_{x \to 0} {{ \sqrt{1+{3 x} + x^2} - \sqrt{1+{3 x} - x^2}} \over x^2} = 1$.

To compute the first limit, let $f_0(x) = \ln(\frac{1+x}{1 + 2x})$ and note that $\lim_{x \to 0} {1 \over x} \ln(\frac{1+x}{1 + 2x}) = f_0'(0) = -1$.

To compute the latter limit, let $f_1(x) = \sqrt{1+{3 x} + x^2}-\sqrt{1+{3 x} - x^2}$, we have $f_1'(0) = 0, f_1''(0) = 2$, hence $\lim_{x \to 0} {{ \sqrt{1+{3 x} + x^2} - \sqrt{1+{3 x} - x^2}} \over x^2} = {f_1''(0) \over 2} = 1$.

Hence the limit is $-1$.

copper.hat
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