$\lim_{n\to \infty}\sqrt[n]{3^n+5^n}$
I'm stuck. Tried to apply the squeeze theorem, but found only the $\le$ side, which is $\sqrt[n]{5^n}$, approaching 5.
How do I proceed from there?
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10$5^n\le 3^n+5^n\le2\cdot 5^n$. – David Mitra Nov 09 '15 at 16:51
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1Here are a few more general questions: How to show $\lim_{n \to \infty} \sqrt[n]{a^n+b^n}=\max {a,b}$? (and other posts linked there) or Convergence of $\sqrt[n]{x^n+y^n}$ (for $x, y > 0$) (and other posts linked there.) – Martin Sleziak Jan 28 '18 at 09:13
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Hint: $$\sqrt[n]{3^n+5^n}=5\sqrt[n]{\left(\frac{3}{5}\right)^n+1}$$
Ben Grossmann
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E.H.E
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The answer is in the comment of David Mitra. $5^n\le 3^n+5^n\le 2*5^n$.
If you take limits, you get $5\le \lim\sqrt[n]{3^n+5^n}\le \lim\sqrt[n]{2}*5$. Since $\lim\sqrt[n]{2}=1$, you get that your limit is $5$.
Please upvote David Mitra's comment, not my answer.
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+1 for the attitude. I was writing a comment to all David to post his comment as an answer when you already posted it. But so glad about the acknowledgement – Shailesh Nov 09 '15 at 17:00
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$$\lim_{n\to\infty}\sqrt[n]{3^n+5^n}=$$ $$\lim_{n\to\infty}\left(3^n+5^n\right)^{\frac{1}{n}}=$$ $$\lim_{n\to\infty}\exp\left(\ln\left(\left(3^n+5^n\right)^{\frac{1}{n}}\right)\right)=$$ $$\lim_{n\to\infty}\exp\left(\frac{\ln\left(3^n+5^n\right)}{n}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\ln\left(3^n+5^n\right)}{n}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\ln\left(5^n\right)+\ln\left(1+\left(\frac{3}{5}\right)^n\right)}{n}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\ln\left(5^n\right)}{n}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{n\ln\left(5\right)}{n}\right)=$$ $$\exp\left(\lim_{n\to\infty}\frac{\ln\left(5\right)}{1}\right)=$$ $$\exp\left(\lim_{n\to\infty}\ln\left(5\right)\right)=$$ $$\exp\left(\ln\left(5\right)\right)=e^{\ln\left(5\right)}=5$$
Jan Eerland
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