Let f : R ⟶ R be a differentiable function. If lim x → ∞ f ( x ) x = 1 ,then there exists a sequence ( x n ) such that x n → ∞ as n → ∞ and lim n → ∞ f ′ ( x n ) = 1.We tried to use the Mean Value Theorem. For each n, there exists x n ∈ [ 0 , n ] such that f ′ ( x n ) = f ( n ) − f ( 0 ) n .Hence, lim n → ∞ f ′ ( x n ) = 1. But, we are not sure that x n → ∞

Question

Let   f  :      R    ⟶      R   be a differentiable function. If      lim          x      →      ∞                  f      (      x      )        x    =  1  ,then there exists a sequence   (      x    n    ) such that       x    n    →  ∞ as   n  →  ∞ and      lim          n      →      ∞            f    ′    (      x    n    )  =  1.We tried to use the Mean Value Theorem. For each n, there exists       x    n    ∈  [  0  ,  n  ] such that      f    ′    (      x    n    )  =            f      (      n      )      −      f      (      0      )        n    .Hence,       lim          n      →      ∞            f    ′    (      x    n    )  =  1. But, we are not sure that       x    n    →  ∞

petyelebxu · Accepted Answer

You can apply the mean-value theorem to the intervals   [  n  ,  2  n  ]:      f    ′    (      x    n    )  =            f      (      2      n      )      −      f      (      n      )              2      n      −      n        =  2            f      (      2      n      )              2      n        −            f      (      n      )        n    →  2  −  1  =  1and       x    n    ≥  n  →  ∞

If lim_(x -> oo) (f(x))/(x)=1, then EE x_n -> oo such that lim_(n -> oo) f′(x_n)=1

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