# Is there an easy way to calculate $\displaystyle\lim_{k \to \infty} \frac{(k+1)^5(2^k+3^k)}{k^5(2^{k+1} + 3^{k+1})}$?

Is there an easy way to calculate $$\lim_{k \to \infty} \frac{(k+1)^5(2^k+3^k)}{k^5(2^{k+1} + 3^{k+1})}$$

Without using L'Hôpital's rule 5000 times?

Thanks!

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Are you sure you can find the limit by applying L'Hôpital's rule 5000 times? – user1551 Mar 11 '13 at 10:40
It should come out to $\frac{1}{3}$, pick out the largest growing terms, I think just having some familiarity with the growth rate of elementary functions is helpful. – Ethan May 16 '13 at 6:14

Hint: Note that $\frac{(k+1)^5}{k^5}\sim 1$ when $k\to\pm\infty$. Now think about the other terms. In fact, think about: $$3^k\left(\left(\frac{2}{3}\right)^k+1\right)/3^{k+1}\left(\left(\frac{2}{3}\right)^{k+1}+1\right)$$

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$x$ should be $k$ – Harmen Mar 11 '13 at 14:41
@Harmen: Thanks for noting me that – Babak S. Mar 11 '13 at 14:44
+++ Very nice! Good Day, dear friend! ;-) – amWhy Mar 11 '13 at 15:52
@amWhy: Nice day to you, dear. – Babak S. Mar 11 '13 at 16:01
Congrats Babak very nice answer – Adi Dani Mar 11 '13 at 16:16

It is a product of the following two expressions

$\frac{(k+1)^5}{k^5}=\left(1+\frac1k\right)^5$

$\frac{2^k+3^k}{2^{k+1}+3^{k+1}}= \frac{3^k(1+(2/3)^k)}{3^{k+1}(1+(2/3)^{k+1})}= \frac13\cdot$ $\frac{1+(2/3)^k}{1+(2/3)^{k+1}}$

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