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Question Number 203909 by Davidtim last updated on 01/Feb/24
prove that 0^n =0 , n>0
provethat0n=0,n>0
Answered by Mathspace last updated on 01/Feb/24
0^n =lim_(x→0^+ ) x^n =lim_(x→0^+ ) e^(nlogx) but lim_(x→0^+ ) nlogx =−∞ (n>0) ⇒0^n =e^(−∞) =0
0n=limx→0+xn=limx→0+enlogxbutlimx→0+nlogx=−∞(n>0)⇒0n=e−∞=0
Answered by witcher3 last updated on 02/Feb/24
0^n =0;n=1 0^n =0.0^(n−1) =0;∀n≥2 for n≥2;n−1≥1⇒0^(n−1) is defined 0.0^(n−1) =0 ⇒∀n≥1 0^n =0
0n=0;n=10n=0.0n−1=0;∀n⩾2forn⩾2;n−1⩾1⇒0n−1isdefined0.0n−1=0⇒∀n⩾10n=0

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