{X/X=4n-1,n e N}write in rule method
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Explanation:
Limit of (1+x/n)n when n tends to infinity [duplicate]
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About lim(1+xn)n 9 answers
Does anyone know the exact proof of this limit result?
limn→∞(1+xn)n=ex
limits proof-writing
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edited Jul 30 '14 at 15:32
Darth Geek
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asked Jul 30 '14 at 15:30
narendra-choudhary
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marked as duplicate by Thomas Andrews, Namaste, Daniel Fischer, user940, M Turgeon Jul 30 '14 at 16:25
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Depends on the definition of ex... – Thomas Andrews Jul 30 '14 at 15:32
have you tried to explicit the power and take the limit? It converge to the taylor series. – Lolman Jul 30 '14 at 15:33
I tried by taking log of both sides, but I don't know what to do after this step. Thought of using L'Hopital's rule. But that ain't helping. – narendra-choudhary Jul 30 '14 at 15:38
You can use this technique. – Mhenni Benghorbal Jul 30 '14 at 15:44
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Using L'Hopital on the log you get:
limn→∞log(1+xn)1/n=limn→∞nn+x−xn2(−n2)=x
– Lolman Jul 30 '14 at 15:46
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A short proof:
(1+xn)n=enlog(1+xn)
Since log(1+x)=x+O(x2) when x→0, we have nlog(1+xn)=x+O(x2n) when n→+∞
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edited Jul 30 '14 at 15:37
Thomas Andrews
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answered Jul 30 '14 at 15:36
Petite Etincelle
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neat, really good – Lolman Jul 30 '14 at 15:37
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Or just use the derivate of log to evaluate limy→0log(1+xy)y, then let y=1/n to get the limit of nlog(1+x/n)... – Thomas Andrews Jul 30 '14 at 15:39
Even more elementary would be to use the basic inequality x/n1+x/n≤log(1+x/n)≤x/n. Then a straightforward application of the squeeze theorem does the trick. – Mark Viola Oct 25 '15 at 16:03
In this case, why does writing (1+xn)n=10nlog10(1+xn) not show that limn→∞(1+xn)n=10x? – Rasputin Jan 19 at 21:17
@Rasputin because the Taylor expansion of log10(1+x) is different – Petite Etincelle Jan 20 at 13:27
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eln(1+xn)n=enln(1+xn)
limn→+∞(1+xn)n=limn→+∞enln(1+xn)=elimn→+∞nln(1+xn)=elimn→+∞ln(1+xn)1n
Apply L'Hopital's Rule:
=elimn→+∞(−xn2)11+xn−1n2=elimn→+∞x1+xn=ex
Therefore,
(1+xn)n→ex