I'm have some doubts about how to correctly solve a complex limit (for me!). I know that I should use kind of asymptotics and the hierarchical order of infinite.

This is the limit, and my idea. I checked with an online calculator and the result is correct, but the calculator applies L'Hospital rule but I can't apply it because I can't use derivative.

$$\lim_{n->\infty} \frac{n+\log{n^2}-2^n}{2(\log{n})^3+n\log{n}+n^2}$$

My idea:

1) For the numerator:

$$\lim_{n->\infty} {n+\log{n^2}-2^n} \sim -1*2^n$$

Considering the hierarchy, $-2^n$ goes first to the infinity, so I can just study $-2^n=-1*2^n$

2) For the denominator, I apply the same idea:

$$\lim_{n->\infty} {2(\log{n})^3+n\log{n}+n^2} = 2(\log^3{n})+nlogn+n^2 \sim n^2$$

So I can study:

$$\lim_{n->\infty} \frac{-1*2^n}{n^2}$$

The numerator goes faster than the denominator to the infinity so answer is $-\infty$

I'm not sure about the entire procedure, even if it sounds realistic to me. Sorry if I wrote something horrible! :)

  • $\begingroup$ could be streamlined a bit, but the idea is fine $\endgroup$ – tired Oct 19 '16 at 8:50
  • $\begingroup$ You mean to say " the numerator goes to infinity faster than the denominator" instead of "the numerator goes first to infinity", right ? $\endgroup$ – Ahmad Bazzi Oct 19 '16 at 8:55
  • $\begingroup$ @ElBazzi Yes, you are right! Faster....wrong translation! $\endgroup$ – LPs Oct 19 '16 at 9:07

To make it a little more rigourous, you can factor out the dominant terms that youd identified, and

$$\lim_{n\to\infty} \frac{n+\log{n^2}-2^n}{2(\log{n})^3+n\log{n}+n^2}= $$ $$\lim_{n\to\infty} \frac{2^n}{n^2}\frac{n2^{-n}+\log{n^2}2^{-n}-1}{2\dfrac{(\log{n})^3}{n^2}+\dfrac{\log{n}}n+1}.$$

By the ratio test,

$$n^k2^{-n}\text{ and }\log^k\,2^{-n}$$ tend to $0$, as $\left(\dfrac {n+1}n\right)^k2^{-(n+1)+n}$ and $\left(\dfrac{\log(n+1)}{\log n}\right)^k2^{-(n+1)+n}$ tend to $1/2$ for all $k>0$

and this implies, with $m=\log_2n=\log n\log_2e$, that

$$\frac{\log^qn}{n^p}=\log_2^qe\left(m^{q/p}2^{-m}\right)^p$$ also tends to $0$ for $p,q>0$.

Now the limit of the large fraction on the right is $\dfrac{0+0-1}{0+0+1}$ and by the previous statements, the requested limit diverges to infinity.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.