3
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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.

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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

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

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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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