Using L'Hopital's RuleDate: 01/22/99 at 05:17:12 From: Mirza Subject: What is the limit of .... Could you give me the proof that e^(-1/3) is the limit for y = (arctan(x)/x)^(1/(x^2)) when x -> 0? Date: 01/22/99 at 08:48:31 From: Doctor Anthony Subject: Re: What is the limit of .... Take logs: ln(y) = (1/x^2)ln[tan^(-1)(x)/x] ln[tan^(-1)(x)/x] ln(y) = ----------------- x^2 Note that tan^(-1)(x)/x -> 1 as x -> 0 and ln(1) = 0. So the righthand side tends to the value 0/0 as x -> 0, and we can apply l'Hopital's rule. Differentiating the top and bottom lines separately we have: [x/tan^(-1)(x)][x/(1+x^2)-tan^(-1)(x)]/x^2 --------------------------------------------- 2x x - tan^(-1)(x) (1+x^2) = ------------------------ 2x^2 tan^(-1)(x) (1+x^2) As x -> 0, this is still 0/0, so we repeat the process. Differentiating the top and bottom separately 1 - [(1+x^2)/(1+x^2) + tan^(-1)(x) 2x] -------------------------------------------- 2(x^2+x^4)/(1+x^2) + 2 tan^(-1)(x) (2x+4x^3) which simplifies to -tan^(-1)(x) ------------------------- x + 2(1+2x^2)tan^(-1)(x) This is still 0/0 so repeat the process. Differentiate: -1/(1+x^2) -------------------------------------- 1 + 2(1+2x^2)/(1+x^2) + 8x tan^(-1)(x) Then multiply top and bottom by (1+x^2): -1 ------------------------------------------ 1 + x^2 + 2(1+2x^2) + 8x(1+x^2)tan^(-1)(x) Putting x = 0, this reduces to: -1 -1 ----- = --- 1 + 2 3 So the limit of ln(y) = -1/3. Thus the limit of y = e^(-1/3). - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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