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We know that $\lim\limits_{n\to\infty}n^{1/n} = 1$. Using this, how can we prove that $\lim\limits_{n\to\infty} (n+1)^{1/n} = 1$?

Recalling the proof of the former limit, I was able to modify it to prove the latter limit. But I was wondering if we could just use the limit we have already calculated to prove this limit which is related to it.

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$1\le(n+1)^{1/n}\le (2n)^{1/n}=2^{1/n}n^{1/n}$. Now Squeeze... –  David Mitra Aug 18 '13 at 15:22
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@DavidMitra Please make your comment into an answer so that I can upvote and accept. –  Maverick Aug 18 '13 at 16:51

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up vote 10 down vote accepted

Note $$1\le (n+1)^{1/n}\le (2n)^{1/n}=2^{1/n}n^{1/n}.$$ Now you could apply the Squeeze Theorem using your known limit.

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HINT: $$\lim_{n\to\infty} (n+1)^{\frac1n}=\left(\lim_{n\to\infty} (n+1)^{\frac1{n+1}}\right)^{\lim_{n\to\infty}\frac{n+1}n}$$

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Could you please elaborate how do you get that limit in the power? –  Maverick Aug 18 '13 at 15:25
    
@Maverick, for positive real value $a$, we have $a^b=(a^c)^{\frac bc}$ where $b,c$ are real numbers proofwiki.org/wiki/Exponent_Combination_Laws/Power_of_Power –  lab bhattacharjee Aug 18 '13 at 15:30
    
That is ok, but why the limit? –  Maverick Aug 18 '13 at 16:44
    
@Maverick, you can use the formula you mentioned –  lab bhattacharjee Aug 18 '13 at 17:13
    
I think what you intended to write was $\left(\lim (n+1)^{1/(n+1)}\right)^\frac{n+1}{n}$ –  Maverick Aug 18 '13 at 17:30

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