I have the following result: lim n → ∞ ( 1 − x n ) n = e − x But how do I get constant lower bounds like ( 1 − 2 x 10 n ) n 2 x ≥ 2 − 1 10 or ( 1 − 2 x 10 n ) n 2 ( x − 1 ) ≥ 4 − 1 5

Question

I have the following result:       lim          n      →      ∞        (  1  −      x    n        )    n    =      e          −      x      But how do I get constant lower bounds like            (      1      −                        2          x                          10          n                    )                                n                      2            x                                ≥      2          −                        1          10                    or            (      1      −                        2          x                          10          n                    )                                n                      2                          (              x              −              1              )                                            ≥      4          −                        1          5

Samuel Hooper · Accepted Answer

I have found the following estimation:      4          −      x        ≤  1  −  x  ≤      2          −      x      where the first inequality hold for   0  &amp;lt;  x  ≤      1    2   and the second for   0  &amp;lt;  x  &amp;lt;  1.With some basic arithmetic rules we can get the bounds asked for in the question.(For others it might also be helpful if someone could provide a proof)

Jonas Huff · Accepted Answer

You can find the minimum value of the function that is on the left in your expression            (      1      −                        2          x                          10          n                    )                                n                      2            x                                ≥      2          −                        1          10                    Remember that to find the minimum value only you have that to find the derivate and to equal it to 0

I have the following result: lim n->infty(1−x/n)^n=e^−x. But how do I get constant lower bounds like (1−2^x/10n)^(n/2^x)>=2^−1/10 or (1−2^x/10n)^(n/2^(x−1))>4^−1/5

Answered question

Answer & Explanation

New Questions in Calculus 1