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To show: $ \lim_{n \rightarrow \infty} 2n ( a^{\frac{1}{2n}} -1)=log(a) $ for a>0.

By definiton: $$ e^{x}=\lim_{m \rightarrow \infty}(1+ \frac{x}{xm} )^{xm} =: a $$ Now take the log of both sides: $$ x= log(\lim_{m \rightarrow \infty}(1+ \frac{x}{xm} )^{xm})=log(a) $$ Substitute xm with 2n, swap lim and log?: $$ x= \lim_{n \rightarrow \infty}log((1+ \frac{x}{2n} )^{2n})=log(a) $$ Using the log rules leads us to this: $$ x= \lim_{n \rightarrow \infty}2n (log(2n+x)-log(2n))=log(a) $$ Well, now it does look pretty similar but still not same as it has to be shown. Any advices? Thanks!

user3342072
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Note that for $|x|<1$ we have

$$ (1+x) \leqslant e^{x}\leqslant \frac{1}{1 - x}\tag{1} $$

Proof : One does not need calculus to prove this inequality; it is a simple corollary of inequality $(1)$ in @Dr.MV's answer here, proved by "non-calculus" means.

Consequently, for $\alpha:=\log a$, $(1)$ allows us to consider the form of the inequality $$ \alpha \leqslant \frac{e^{\alpha (1/2n)}-1}{1/2n}\leqslant \frac{\alpha}{1-\alpha (1/2n)} $$ from which the squeeze theorem implies $$ \lim_{n\to\infty}\frac{e^{\alpha (1/2n)}-1}{1/2n}=\alpha. $$

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ki3i
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  • Thank you, sir, for your efforts it's very plausible proof you have, but integrals as well as derivatives were not introduced yet, therefore we are not allowed to use this handy tools yet. – user3342072 Jan 10 '16 at 16:55