As written in Wiki, $e^x=\lim_{n \to \infty}\left(1+\frac{x}{n}\right)^n$.
However, does anyone agree that $$e^x = \lim_{n \to \infty}\left(\frac{1+0.5\frac{x}{n}}{1-0.5\frac{x}{n}}\right)^n ?$$
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@Travis: Seriously? [left]? – Asaf Karagila Sep 17 '14 at 20:05
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@AsafKaragila Just a typo, I'm not sure how that got into the tag list. – Travis Willse Sep 18 '14 at 01:54
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You can push the limit operation to the numerator and the denominator and evaluate the limits separately. The result will be $\displaystyle\frac{e^{0.5x}}{e^{-0.5x}} = e^x$.
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so it seem i be using mid point method does hold as well. using y1+((y1+y2)/2)(x/n)=y2 and continue sequence as y2+((y2+y3)/2)(x/n)=y3, continue the sequence and reach the above formula – chuackt Sep 15 '14 at 16:32
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Yes. Your expression is correct.
Lucian
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Hint: Write $\bigg(1+0.5\dfrac xn\bigg)=\bigg(1-0.5\dfrac xn\bigg)+2\cdot0.5\dfrac xn$ – Lucian Sep 15 '14 at 11:31
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You are right:
$$\lim _{n\rightarrow \infty } \left( \left( 1+{\frac {ax}{n}} \right) \left( 1+{\frac {bx}{n}} \right) ^{-1} \right) ^{n}={{\rm e}^{x \left( a-b \right) }} $$
$$\lim _{n\rightarrow \infty } \left( \left( 1+{\frac {ax}{n}} \right) \left( 1+{\frac { \left( a-1 \right) x}{n}} \right) ^{-1} \right) ^{n }={{\rm e}^{x}} $$
Juan Ospina
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