Wikipedia:Reference desk/Archives/Mathematics/2019 February 1

= February 1 =

Limits
I managed to find a page online that told me that lim n->inf (1-1/n)^n is 1/e but it doesn't have an explanation. Also I'd like to read more about this kind of limit. Can someone point me to some material on this? Thank you. RJFJR (talk) 18:16, 1 February 2019 (UTC)
 * See the last formula in Exponential function. The limit is an example of an Indeterminate form, specifically 1∞. As such it can be solved be applying L'Hôpital's rule. --RDBury (talk) 22:48, 1 February 2019 (UTC)
 * It may be circular logic to use L'Hôpital's rule as the calculus itself is based on the above limit in part. Ruslik_ Zero 08:25, 2 February 2019 (UTC)
 * True, you wouldn't use L'Hôpital's rule to prove the limit if you were trying to prove derivative formula's from scratch. But once you've established derivative formulas then L'Hôpital is fair game and probably the easiest method for evaluating an indeterminate form, especially if you didn't bother keeping track of all the ingredients that were used to prove those derivative formulas. The same sort of issue comes up with derivatives of trig functions and I've had more than one frustrated Calc I student who didn't understand why they had to learn the squeeze theorem to find the limit of sin x/x when it's so much easier to use L'Hôpital. In any case, the OP just asked for some links, not proofs, and I think L'Hôpital is still relevant. --RDBury (talk) 12:20, 2 February 2019 (UTC)

If you're familiar with the Taylor series for exp(x), you know $$e^x=1+x+{x^2\over{2!}}+\cdots$$ and if x is near zero, you can ignore the higher order terms. So you get $$e^x\approx 1+x$$ when x is small. Substituting $$x=-1/n$$, that means $$e^{-1/n}\approx 1-{1/n}.$$ Raising both sides to the nth power, you get $$e^{-1}\approx \left(1-1/n\right)^n$$ q.e.d. 173.228.123.166 (talk) 10:41, 2 February 2019 (UTC)
 * you can ignore the higher order terms is faulty logic. The correct calculation needs to take care of the residual: $$e^x\approx 1+x+O(x^2)$$ (see big O notation), hence $$e^{-\frac{1}{n}}\approx 1-\frac{1}{n}+O(\frac{1}{n^2})$$, hence $$\left(e^{-\frac{1}{n}}\right)^n\approx \left(1-\frac{1}{n}+O(\frac{1}{n^2})\right)^n $$. The right hand side of that latter equation is equivalent to $$\left(1-\frac{1}{n}\right)^n +O(\frac{1}{n}) $$ where the residual goes to zero as n grows large, which means we are in the clear, but it is not always necessarily the case when manipulating Taylor series.
 * For an example of what the faulty logic may lead you, consider $$\frac{e^{2x}-2x-1}{x^2}$$. Its limit as x goes to 0 is 1, but if you had truncated the Taylor series to $$e^{2x}\approx 1+2x$$ "because we can ignore the higher order terms", you would have erroneously deduced the limit was 0. Making the computation with the residual yields $$\frac{e^{2x}-2x-1}{x^2} = \frac{1+2x+O(x^2)-2x-1}{x^2} = O(1)$$, and the fact that the residual does not tend to 0 as x goes to 0 in the end results shows you need more terms in the Taylor series.  Tigraan Click here to contact me 14:03, 2 February 2019 (UTC)

Thank you everybody. (I never paid enough attention to those 1^inf indeterminate forms. I'll go study up on it now.  And the calculations using big-O notation is great to see.) RJFJR (talk) 00:38, 3 February 2019 (UTC)