Rooting For The Limit (Posted on 2008-04-23)

	Submitted by K Sengupta
Rating: 3.5000 (2 votes)
Solution:	(Hide)
Let M = Limit ((2y + 22y + 23y)/3)1/y as y → 0 or, ln M = Limit ((ln (2y + 22y + 23y) - ln 3)/y as y → 0 Since the expression in the rhs is of the form 0/0, differentiating both the numerator and the denominator in terms of L'Hospital's rule, we obtain: ln M = Limit (2y*ln 2+ 22y*ln 4 + 23y*ln 8)/(2y + 22y + 23y) as y → 0 = (ln 2 + ln 4 + ln 8)/3 = (ln 64)/3 Thus, M = (64)1/3 = 4 Consequently, the required limit is 4. Solution To The Bonus Question: Let N = Limit ((2y + 6y + 18y)/3)1/y as y → 0 or, ln N = Limit ((ln (2y + 6y + 18y) - ln 3)/y as y → 0 Since the expression in the rhs is of the form 0/0, differentiating both the numerator and the denominator in terms of L'Hospital's rule, we obtain: ln N = Limit (2y*ln 2 + 6y*ln 6 + 18y*ln 18)/(2y + 6y + 18y) as y → 0 = (ln 2 + ln 6 + ln 18)/3 = (ln 216)/3 Thus, N = (216)1/3 = 6 Consequently, the required limit is 6. ------------------------------------------ Note: In general, for n positive real numbers x1, x2, ……., xn, it can be similarly shown that: Limit ((x1y+ x2y+.....+ xny)/n)1/y = (x1*x2…..* xn)1/n y → 0

	Subject	Author	Date
	not very rigorous solution	Paul	2008-04-25 23:59:00
	Hint	K Sengupta	2008-04-25 11:41:53
	re: Charlie	hoodat	2008-04-24 18:05:41
	re: Solution	Charlie	2008-04-23 17:23:23
	Solution	hoodat	2008-04-23 14:45:34
	computer exploration (possible spoiler)	Charlie	2008-04-23 11:36:44