What is lim n→∞ (2)^(1/n)? Why is it equal to 1? All proofs shown are about n^(1/n) or something like this form. What about 2?
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$\lim_{n \to \infty }\tfrac1n = 0$ and $2^0=1$. – azif00 Feb 26 '21 at 09:06
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1do you want epsilon proof? – Aditya Dwivedi Feb 26 '21 at 09:06
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1$n >\frac 1 {\log_2(1+\epsilon)} $ implies that $|2^{1/n}-1| <\epsilon$. – Kavi Rama Murthy Feb 26 '21 at 09:09
2 Answers
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$\lim _{n \rightarrow \infty} \frac{1}{n}=0$ So, $\lim _{n \rightarrow \infty} 2^{\frac{1}{n}}=2^{0}=1$ As denominator gets bigger and bigger while numerator being constant then fraction gets closer and closer to zero.
Ashish Kumar
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As the base $b=2$ is a constant,
$\lim_{n\rightarrow\infty} 2^{(1/n)} = 2^{\lim_{n\rightarrow\infty} (1/n)} = 2^0 = 1.$
Wuestenfux
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