imported_Lucent
Member
- Jun 26, 2006
- 40
- 0
- 0
Eeezee said: Someone help me with the 0.9 repeating = 1 proof
IF 0.9 with nines repeating comes from lim (x->1) x=1 and the function f(x)=x THEN you can prove using theorems in calculus that f(x) at x=1 APPROACHES AND BECOMES EQUAL TO 1 (basically the limit equals the function value and it is a continous function) FOR THE SPECIFIC FUNCTION, LIMIT AND INPUT VALUE. BUT THE LIMIT DOES NOT NECESSARILY EQUAL THE VALUE OF THE FUNCTION HENCE THERE IS NO PROOF THAT 0.9 WITH NINES REPEATING EQUALS 1. HERE IS A EXAMPLE:
f(x)= { x if x is not equal to one
50000000 if x is equal to one}
Another example:
lim (x->1) (x-1)/(x^2-1)=0.5
(x-1)/(x^2-1) at x=1 is INDETERMINATE. FUNCTION VALUE DOES NOT EQUAL THE LIMIT!
IF 0.9 with nines repeating comes from lim (x->1) x=1 and the function f(x)=x THEN you can prove using theorems in calculus that f(x) at x=1 APPROACHES AND BECOMES EQUAL TO 1 (basically the limit equals the function value and it is a continous function) FOR THE SPECIFIC FUNCTION, LIMIT AND INPUT VALUE. BUT THE LIMIT DOES NOT NECESSARILY EQUAL THE VALUE OF THE FUNCTION HENCE THERE IS NO PROOF THAT 0.9 WITH NINES REPEATING EQUALS 1. HERE IS A EXAMPLE:
f(x)= { x if x is not equal to one
50000000 if x is equal to one}
Another example:
lim (x->1) (x-1)/(x^2-1)=0.5
(x-1)/(x^2-1) at x=1 is INDETERMINATE. FUNCTION VALUE DOES NOT EQUAL THE LIMIT!