Have I made a new limit that tends to e?I have made two formulae. One tending to e and one tending to 3: e = lim n → ∞ ( 2 × 10 n − 1 2 × 10 n − 3 ) 10 n − 1 3 = lim n → ( − ∞ ) ( 2 × 10 n − 1 2 × 10 n − 3 ) 10 n − 1 Is there a way of proving that these formulae are indeed true? Thanks.

Question

Have I made a new limit that tends to   e?I have made two formulae. One tending to   e and one tending to   3:  e  =      lim          n      →      ∞                  (                        2          ×                      10            n                    −          1                          2          ×                      10            n                    −          3                    )                      10        n            −      1        3  =      lim          n      →      (      −      ∞      )                  (                        2          ×                      10            n                    −          1                          2          ×                      10            n                    −          3                    )                      10        n            −      1      Is there a way of proving that these formulae are indeed true? Thanks.

driogairea1 · Accepted Answer

The second one is true because       lim          n      →      (      −      ∞      )            10    n    =  0, so we get      lim          n      →      (      −      ∞      )                  (                  2      ×              10        n            −      1              2      ×              10        n            −      3                          )                            10        n            −      1        =            (                  −      1              −      3                          )                    −      1        =  3.For the first one, note that      lim          m      →      ∞                  (                  2      m      +      1              2      m      −      1                          )              m    =      lim          m      →      ∞                  (        1  +      1          m      −      1              /            2                          )                    m      −      1              /            2                  (        1  +      1          m      −      1              /            2                          )                    1              /            2        .The second bracket goes to   1 and the first to   e. Since your sequence is just a subsequence of this one, it has the same limit.

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