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I was wondering if we can solve this limit without using squeeze (sandwich) theorem. $$\lim_{n\to \infty}(3^n+5^n)^{2/n}$$

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  • $\begingroup$ Use the fact that $3^n\ll5^n$. $\endgroup$ – Edward Jiang Dec 29 '14 at 21:38
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    $\begingroup$ @Shrey what exactly are you looking for ? The first principle needs $|(3^n+5^n)^{2/n} - 25|$ to be squeezed to $0$ ;) $\endgroup$ – r9m Dec 29 '14 at 21:41
  • $\begingroup$ Why would you want to? All of the below proofs seem more complicated and less direct than the observation $5^n \leq 3^n + 5^n \leq (2)5^n$. $\endgroup$ – Michael Dec 29 '14 at 21:59
  • $\begingroup$ Thank you all for responding. I got what I was looking for in solutions below. I started doing this using l'Hospital's but got stuck. Although applying squeeze theorem is easier but it doesn't hurt to know alternative solution. $\endgroup$ – Shrey Dec 29 '14 at 22:21
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    $\begingroup$ Yep. Just post it on math.SE. . . oh I get it now. $\endgroup$ – meawoppl Dec 30 '14 at 1:18
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Certainly. $$ \lim_{n \to +\infty} \exp\left(\ln\left((3^n + 5^n)^{2/n}\right)\right) = \lim_{n \to +\infty} \exp \left(\frac{2\ln(3^n + 5^n)}{n}\right) = \exp\left(\lim_{n \to +\infty}\frac{2\ln(3^n + 5^n)}{n}\right) $$

Now, let's solve $\displaystyle \lim_{n \to +\infty}\frac{2\ln(3^n + 5^n)}{n}$. We have $\displaystyle \frac{+\infty}{+\infty}$, so we can apply l'Hospital's rule straight away, giving us: $$ \lim_{n \to +\infty}\frac{2 \ln(3) 3^n + 2 \ln(5) 5^n}{3^n + 5^n} = 2 \ln(5) $$

So, the limit is $\displaystyle \exp(2 \ln(5)) = 25$.

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$$\lim_{n \to \infty} f(x) = e^{\lim_{n \to \infty}{\ln(f(x))}}=e^{\lim_{n \to \infty}{\frac2n\ln(3^n+5^n)}}=e^{2\lim_{n \to \infty}{\frac{\ln{3^n+5^n}}{n}}}$$

Now let's use a L'Hopital's rule:

$$e^{2\lim_{n \to \infty}{\frac{\ln{3^n+5^n}}{n}}}=e^{2\lim_{n \to \infty}{\frac{3^n\ln3+5^n\ln5}{3^n+5^n}}}=e^{2\lim_{n \to \infty}{\ln5+\frac{3^n\ln3-3^n\ln5}{3^n+5^n}}}=e^{2\ln5}=e^{\ln25}=25$$

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$$\begin{align} &\lim_{n\to \infty}(3^n+5^n)^{2/n}\\ =&\lim_{n\to \infty}\exp\left( {2\log(3^n+5^n)/n}\right)\\ =&\exp\left({\lim_{n\to \infty}{2\log(3^n+5^n)/n}}\right)\\ =&\exp\left({\lim_{n\to \infty}{\frac{2\log(3)3^n+2\log(5)5^n}{3^n+5^n}}}\right)\\ =&\exp\left({\lim_{n\to \infty}{\frac{2\log(3)(3/5)^n+2\log(5)}{(3/5)^n+1}}}\right)\\ =&\exp\left({{\frac{[0]+2\log(5)}{[0]+1}}}\right)\\ =&\exp\left(2\log 5\right)\\ =&25\\ \end{align}$$

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    $\begingroup$ Explanations: 1.Limit 2.$e^{\log(x)}=x$ 3. $\lim a^{f(x)}=a^{\lim f(x)}$ 4. L'Hopital 5. Divide by $5^N$ 6. Evaluate the limit 7. Simplify 8. Simplify $\endgroup$ – Cyclohexanol. Dec 29 '14 at 21:44
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You may just write, as $n$ tends to $+\infty$,

$$ \begin{align} (3^n+5^n)^{2/n}&=e^{\frac2n\log \left(3^n+5^n \right)}\\\\ &=e^{\frac2n\log \left(5^n\right)+ \frac2n\log \left(1+(3/5)^n \right)}\\\\ &=e^{2\log 5+ \frac2n \log \left(1+(3/5)^n \right)}\\\\ &=e^{2\log 5+ \frac2n (3/5)^n }\\\\ &\sim e^{2\log 5}\times e^0\\\\ &\sim25 \end{align} $$ where we have used $$ \log (1+x)\sim_0 x $$ with $x:=(3/5)^n$, $n$ being great.

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  • $\begingroup$ Question: why can he write as $n$ tends to $+\infty$? Why can we disregard $-\infty$? $\endgroup$ – Newb Dec 29 '14 at 22:59
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$$\lim_{n \to \infty }\sqrt[n]{(3^n+5^n)^2}=\\\lim_{n \to \infty }\sqrt[n]{(5^n((\frac{3}{5})^n+1))^2}=\\5^2\lim_{n \to \infty }\sqrt[n]{((\frac{3}{5})^n+1)^2}=\\$$as we now $$\lim_{n \to \infty }(\frac{3}{5})^n=0$$so $$5^2\lim_{n \to \infty }\sqrt[n]{((\frac{3}{5})^n+1)^2}=5^2*\sqrt[n]{(0+1)^2}=25$$

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    $\begingroup$ This assumes without proof that $\lim_{n\rightarrow\infty} f(n)^{1/n} = 1$ whenever $\lim_{n\rightarrow\infty} f(n) = 1$ (true, but this is a significant step). $\endgroup$ – Michael Dec 29 '14 at 21:52

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