2007-02-22 08:35:30 · 7 answers · asked by david c 1 in Science & Mathematics ➔ Mathematics
It doesn't. 1+x approaches infinity as x approaches infinity. 1/(infinity) approaches zero.
2007-02-22 08:42:19 · answer #1 · answered by Anonymous · 2⤊ 0⤋
It doesn't. Your expression is not correct. As x tends to infinity (gets very very big) then 1 + x tends to infinity. So you have a sequence of fractions in which 1 is being divided by larger and larger numbers. You can convince yourself of this by considering the term when x = 1,2,4,10 etc. You get:
1/2, 1/3, 1/5, 1/10 ....
You can see that as x gets bigger and bigger 1/(1+x) gets smaller and smaller. So 1/(1 + x) tends to zero as x tends to infinity.
If it were the other way round, that is x tends to zero, then
1/(1 + x) tends to one.
2007-02-22 19:25:53 · answer #2 · answered by Anonymous · 0⤊ 0⤋
It tends to 0, not 1. The numerator stays at 1, yet the denominator 1+x grows to infinity, so we have 1/∞ = 0.
2007-02-22 08:43:32 · answer #3 · answered by Scythian1950 7 · 0⤊ 0⤋
1/(1+x) tends to zero as x goes to infinity.
2007-02-22 08:41:51 · answer #4 · answered by jason 2 · 1⤊ 0⤋
1/(1 + x)
D = R -(-1) = {x elements of R | x must be different of -1}. In the fractions, the denominator cannot be equal to zero. If zero, fraction becomes null.
><
2007-02-24 17:29:53 · answer #5 · answered by aeiou 7 · 0⤊ 0⤋
I'm afraid that your initial statement is incorrect. It tends to 0. This is because you have a fraction where the bottom is getter increasingly bigger so the whole thing is getting increasingly smaller.
2007-02-22 08:43:29 · answer #6 · answered by Anonymous · 0⤊ 0⤋
no, it tends to 0.
2007-02-22 08:42:31 · answer #7 · answered by Anonymous · 0⤊ 0⤋