Prove that the limit of 1/x is zero as x approaches infinity.?

Question

It is well known that the limit of 1/x is zero as x approaches infinity, but too many don't know how to prove it. Using the definition of limits, how to prove? i could not understand the book proof.

Answer 1

Definition of the statement lim (x->∞) 1/x = 0 is:
For any ε > 0 there exists an N > 0 such that if x > N, then |1/x - 0| < ε.

So let ε > 0 and let N = 1/ε. Then if x > N, 0 < 1/x < 1/N = ε, so |1/x - 0| < ε.

So lim (x->∞) 1/x = 0.

Answer 2

it fairly is the two 3 solutions. - If the decrease merely techniques 0 then the decrease of a million/x won't exist. - If the decrease techniques 0 from the remarkable section that's lim (x->0-) then x <0 -> u can attempt plugging in a sort smaller than 0-> f(a million/x) for any x<0 would be a unfavourable type. So decrease of a million/x as x techniques 0- would be unfavourable INFINITY. -If the decrease techniques 0 from the left section that's lim(x->0+) then x>0 => u can attempt plugging in a a sort extra advantageous than 0 -> f(a million/x) for any x>0 would be an excellent type. So decrease of a million/x as x techniques 0+ would be valuable INFINITY. P/s : I study the remark and observed the question grew to become into from the remarkable section that's lim(x->0+) so the respond could be valuable infinity.

Answer 3

I don't know of its formal proof but here's how I would prove it:

Infinity = 1/0

lim 1/x (as x --> Infinity) = 1/Infinity = 1/(1/0) = 1 * (0/1) = 0