How do I prove that the limit as one function goes to infinity is equal to another function?I was playing around with the integral ∫ x n d x and noticed that it is always x n − 1 n − 1 + c except for the singular case n = − 1. So I could pick n arbitrarily close to − 1 and the formula works, but as soon as I hit − 1 it breaks (so to speak)!Here's my work thus far: ∫ x − ( n + 1 ) / n d x = − n x − 1 / n + cWorking with n = 100 with the initial condition y ( 1 ) = 0 I got: − 100 x − 1 / 100 + c = 0 − 100 + c = 0 c = 100So I figure the following could be true: − n x − 1 / n + n ≈ ln ⁡ xHow can I prove that as I take n → ∞ the two functions are equivalent? lim n → ∞ − n x − 1 / n + n = ln ⁡ xOr equivalently, e lim n → ∞ − n x − 1 / n + n = x

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How do I prove that the limit as one function goes to infinity is equal to another function?I was playing around with the integral   ∫      x    n    d  x and noticed that it is always             x              n        −        1                    n      −      1        +  c except for the singular case   n  =  −  1. So I could pick n arbitrarily close to   −  1 and the formula works, but as soon as I hit   −  1 it breaks (so to speak)!Here&#039;s my work thus far:  ∫      x          −      (      n      +      1      )              /            n        d  x  =  −  n      x          −      1              /            n        +  cWorking with   n  =  100  with the initial condition   y  (  1  )  =  0 I got:  −  100      x          −      1              /            100        +  c  =  0  −  100  +  c  =  0  c  =  100So I figure the following could be true:  −  n      x          −      1              /            n        +  n  ≈  ln  ⁡  xHow can I prove that as I take   n  →  ∞ the two functions are equivalent?      lim          n      →      ∞        −  n      x          −      1              /            n        +  n  =  ln  ⁡  xOr equivalently,      e                  lim                  n          →          ∞                    −      n              x                  −          1                      /                    n                    +      n        =  x

Haleigh Vega · Accepted Answer

After re-writing, you get  −  n      x          −      1              /            n        +  n  =  n  (  1  −      e          −      1              /            n      ln      ⁡      x        )  .Now you can use Taylor expansions, for instance, to get  n  (  1  −      e          −      1              /            n      ln      ⁡      x        )  =  n  (  1  −  (  1  −      1    n      ln  ⁡  x  +  O  (      n          −      2        )  )  =  ln  ⁡  x  +  O  (      n          −      1        )  .Then      lim          n      →      ∞        n  −  n      x          −      1              /            n        =  ln  ⁡  x  .

wanaopatays · Accepted Answer

The key to this is to be careful with the integral, and in particular, the limits of integration. Note that      ∫    1    t        x    n    =                    t                  n          +          1                    −      1              n      +      1        ,  n  ≠  1So now we want to compute      lim          n      →      −      1                          t                  n          +          1                    −      1              n      +      1        .By making a change of variables,   n  ↦  n  −  1 , we get      lim          n      →      0                          t                  n                    −      1        n  which is known to be the natural logarithm.A similar type of thing happens with the integral      lim          n      →      1        ∫  sin  ⁡  (  n  x  )  sin  ⁡  (  x  )  d  x  .

How do I prove that the limit as one function goes to infinity is equal to another function? I was

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