I want to prove $$\lim_{x\to 0} \frac{e - (1+x)^\frac{1}{x}}{x}=\frac{e}{2}$$ without using L'Hôpital's rule, and Taylor Series.
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i want dont use Taylor series or [http://strg.xzn.ir/pictures/9_n.jpg] – DnShVr Aug 01 '14 at 08:05
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1I inderstand that. However, I asked what you DID try, not what you don't want to try. – 5xum Aug 01 '14 at 08:07
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If you do not want Taylor series, precise it in the post. – Claude Leibovici Aug 01 '14 at 08:23
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Maybe you should give a a list what is allowed, and why you do not want some standard methods. – gammatester Aug 01 '14 at 08:39
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i want proof this directly, likely proof of $\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$ – DnShVr Aug 01 '14 at 08:55
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Hint
First, consider $A=(1+x)^\frac{1}{x}$. Take the logarithms of both sides so $$\log(A)=\frac{1}{x} \log(1+x)$$ Use the Taylor series of $\log(1+x)$; multiply the result by $\frac{1}{x}$; take the exponential; use the Taylor series and so on.
I am sure that you can take from here.
Claude Leibovici
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@8pir. Not at all. I think (but I can be wrong) that this is a very simple step-by-step method. – Claude Leibovici Aug 01 '14 at 08:01
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1Taking the Taylor series when L'Hospital should not be used is... I don't know, it feels like cheating... – 5xum Aug 01 '14 at 08:08
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@5xum. Your point is interesting indeed ! So, I confess that I don't know !! But the OP did not write "without using derivatives", isn't ? Any suggestion from your side (I hate to feel that I could be cheating). Cheers and thanks :-) – Claude Leibovici Aug 01 '14 at 08:21
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yes, i want proof without using derivatives. i want proof this directly, likely proof of $\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$ – DnShVr Aug 01 '14 at 09:00
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You know that $\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$ but this is not sufficient for your problem. What you need is $(1+x)^{\frac{1}{x}}=e + something$ and I do not see how you could get $something$ without using Taylor or derivatives. – Claude Leibovici Aug 02 '14 at 05:07
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There is a direct proof but prepare, it is not going to be nice. We are going to use the following well-known limits: $$\lim_{x \to 0}{\frac{e^x-1}{x}}=1$$ $$\lim_{x \to 0}{\frac{\ln{(x+1)}}{x}}=1$$ If you want a direct proof of them you can easily find it online.
Now let's go to the real deal; the limit you are asking for.
$$\lim_{x \to 0}{\frac{e-(1+x)^{\frac{1}{x}}}{x}}$$ $$=\lim_{x \to 0}{\frac{e-e^{\frac{\ln{(1+x)}}{x}}}{x}}$$ $$=\lim_{x \to 0}{\frac{e^{\frac{\ln{(1+x)}}{x}}-e}{-x}}$$ $$=\lim_{x \to 0}{\frac{e(e^{\frac{\ln{(1+x)}}{x}-1}-1)}{-x}}$$ $$=\lim_{x \to 0}{\frac{e(e^{\frac{\ln{(1+x)}}{x}-1}-1)(\frac{\ln{(1+x)}}{x}-1)}{-x(\frac{\ln{(1+x)}}{x}-1)}}$$ $$=\lim_{x \to 0}{\frac{e(\frac{\ln{(1+x)}}{x}-1)}{-x}}$$ $$=e\lim_{x \to 0}{\frac{\frac{\ln{(1+x)}}{x}-1}{-x}}$$ $$=e\lim_{x \to 0}{\frac{x-\ln{(1+x)}}{x^2}}$$ $$=\frac{e}{2}$$
Now let us show that
$$\lim_{x \to 0}{\frac{x-\ln{(1+x)}}{x^2}}=\frac{1}{2}$$
$$L=\lim_{x \to 0}{\frac{x-\ln{(1+x)}}{x^2}}$$ We substitute x=2x and get: $$L=\lim_{x \to 0}{\frac{2x-\ln{(1+2x)}}{4x^2}}$$ $$L=\lim_{x \to 0}{\frac{2(x-\frac{1}{2}\ln{(1+2x))}}{4x^2}}$$ $$2L=\lim_{x \to 0}{\frac{x-\frac{1}{2}\ln{(1+2x)}}{x^2}}$$ Now lets substract the original limit $L$ from this new limit $2L$.
$$2L-L=L=\lim_{x \to 0}{\frac{x-\frac{1}{2}\ln{(1+2x)}-x+\ln{(1+x)}}{x^2}}$$ $$L=\lim_{x \to 0}{\frac{\ln{(1+2x)}-2\ln{(1+x)}}{-2x^2}}$$ $$L=\lim_{x \to 0}{\frac{\ln{(\frac{1+2x}{(1+x)^2})}}{-2x^2}}$$ $$L=\lim_{x \to 0}{\frac{\ln{(1-\frac{x^2}{(1+x)^2})}}{-2x^2}}$$ $$L=\lim_{x \to 0}{\frac{-x^2}{-2x^2(1+x^2)}}=\frac{1}{2}$$
Jesus
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