When doing my test prep, I stumbled upon this particular exercise:

$$\lim_{n \to \infty} \frac{3^{2n + 1} + 2^{3n + 1}}{7^{n+2} + 9^n} $$

I tried to solve it through some algebraical juggling, but without luck. Even trying l'Hospitals rule (which I officially don't know yet) didn't yield a solution. When I inputted the task into Wolfram, I found out the solution is $3$. Even after that, I've not been able to find the steps which are needed to get that solution.

Judging by the value 3, it would seem that I somehow want to "reduce" the fraction so that I get only

$$\lim_{n \to \infty} \frac{3^{2n + 1}}{9^n} $$

from which 3 would easily follow, but I'm not sure how to do that.


For these kind of limits the trick is to factor out the stronger terms form the numerator and the denominator

$$ \frac{3^{2n + 1} + 2^{3n + 1}}{7^{n+2} + 9^n}=\frac{3^{2n + 1}}{9^n}\frac{1 + \frac{2^{3n + 1}}{3^{2n + 1}}}{\frac{7^{n+2}}{9^n}+1}$$

and since $$\frac{1 + \frac{2^{3n + 1}}{3^{2n + 1}}}{\frac{7^{n+2}}{9^n}+1} \to \frac{1+0}{0-1}=1$$

the original limit reduces to evaluate

$$\lim_{n \to \infty} \frac{3^{2n + 1}}{9^n}$$

  • $\begingroup$ Thank you for clarifying how to reduce the fraction into the simpler form. I chose your answer because it more concretely shows the reasoning behind the process. $\endgroup$ – Sh4rP EYE Oct 28 '18 at 13:17

We rewrite the sequence as $$\lim_{n \to \infty} \frac{3\cdot9^n+ 2\cdot8^n}{49\cdot7^n+9^n}$$ and then divide by $9^n$: $$\lim_{n \to \infty} \frac{3+ 2(8/9)^n}{49(7/9)^n+1}$$ Now $(7/9)^n$ and $(8/9)^n$ tend to zero, and the limit thus reduces to $$\frac31=\frac{3\cdot9^n}{9^n}=\frac{3^{2n+1}}{9^n}=3$$ The overall division by $9^n$ makes the intermediate form $\frac{3^{2n+1}}{9^n}$ transparent in the calculations.

  • $\begingroup$ Your solution is correct (+1) but note that the OP was asking in particular how to reduce the calculation to the simpler limit. $\endgroup$ – user Oct 28 '18 at 11:20
  • $\begingroup$ @gimusi just multiply back $3/1$ by $9^n$, then? Dropping off slower-growing terms is only an informal heuristic. $\endgroup$ – Parcly Taxel Oct 28 '18 at 11:21
  • $\begingroup$ As I said, your method is fine. What I'm claiming is that the OP was asking (also) for a way or justification to reduce to the simpler limit. I've just adresed that point. But also your way is effective and can be useful to give a general way to proceed. I also usually proceed in that way! Bye $\endgroup$ – user Oct 28 '18 at 11:26

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