L lookagain Elite Member Joined Aug 22, 2010 Messages 3,250 Jan 29, 2011 #1 I made this problem. Calculate this: \(\displaystyle \lim_ {x \to 0^+}\{1 - [1 - (x^x)]^x \}^x\) \(\displaystyle You \ may \ use \ \lim_{x \to 0^+}(x^x) \ = \ 1.\)
I made this problem. Calculate this: \(\displaystyle \lim_ {x \to 0^+}\{1 - [1 - (x^x)]^x \}^x\) \(\displaystyle You \ may \ use \ \lim_{x \to 0^+}(x^x) \ = \ 1.\)
tkhunny Moderator Staff member Joined Apr 12, 2005 Messages 11,325 Jan 29, 2011 #2 Seems to be screaming for the binomial theorem.
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,250 Jan 30, 2011 #3 lookagain said: \(\displaystyle \lim_ {x \to 0^+}\{1 - [1 - (x^x)]^x \}^x\) \(\displaystyle Let \ y \ belong \ to \ the \ positive \ Reals.\) \(\displaystyle You \ may \ use \ \lim_{y \to 0^+}(y^y) \ = \ 1. ** \ \ \ \ \ Edit \ on \ the \ variable\) Click to expand... Hints: ------- For 0 < x < 1, x^x < 1 . . . . . . Why? As x approaches 0 from the right, (1 - x^x) approaches 0 from above . . . . . Why? \(\displaystyle Then \ \lim_{x \to 0} (1 - x^x)^x \ is \ of \ the \ form \ of **\)
lookagain said: \(\displaystyle \lim_ {x \to 0^+}\{1 - [1 - (x^x)]^x \}^x\) \(\displaystyle Let \ y \ belong \ to \ the \ positive \ Reals.\) \(\displaystyle You \ may \ use \ \lim_{y \to 0^+}(y^y) \ = \ 1. ** \ \ \ \ \ Edit \ on \ the \ variable\) Click to expand... Hints: ------- For 0 < x < 1, x^x < 1 . . . . . . Why? As x approaches 0 from the right, (1 - x^x) approaches 0 from above . . . . . Why? \(\displaystyle Then \ \lim_{x \to 0} (1 - x^x)^x \ is \ of \ the \ form \ of **\)
