Proof ' that ln ⁡ ( x ) convergesWhere is the flaw in the following 'proof '? lim x → ∞ [ d d x { ln ⁡ ( x ) } ] = lim x → ∞ [ 1 x ] = 0 ⟹ lim x → ∞ [ ln ⁡ ( x ) ] = constant ∈ R ,therefore ln ⁡ ( x ) converges to some real number. ◻

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Proof &#039; that   ln  ⁡  (  x  ) convergesWhere is the flaw in the following &#039;proof &#039;?      lim          x      →      ∞            [                  d                              d                x                    {      ln      ⁡      (      x      )      }        ]    =      lim          x      →      ∞            [          1      x        ]    =  0    ⟹        lim          x      →      ∞        [  ln  ⁡  (  x  )  ]  =  constant  ∈      R    ,therefore   ln  ⁡  (  x  ) converges to some real number.   ◻

Franklin Frey · Accepted Answer

It&#039;s not even clear what it means to say that       lim          x      →      ∞        f  (  x  ) &quot;is a constant,&quot; since it isn&#039;t a function of any variable, it&#039;s just a limit, if the limit exists.But consider   f  (  x  )  =      x  . We know that       lim          x      →      ∞            d          d      x        f  (  x  )  =  0, but is       lim          x      →      ∞            x   a constant?One thing we can say, if       lim          x      →      ∞            d          d      x        f  (  x  )  =  0, is that if   D is any constant, then      lim          x      →      ∞            (    f    (    x    +    D    )    −    f    (    x    )    )    =  0This claim is just nonsense:If      lim          x      →      ∞            [                  d                              d                x                    {      ln      ⁡      (      x      )      }        ]    =  0then      lim          x      →      ∞        ln  ⁡  (  x  )  =      lim          x      →      ∞            [    ∫    0          d        x    ]    =  constantThe fundamental theorem of calculus does not even remotely support that. We have:  ln  ⁡  (  x  )  =      ∫          1        x        1    t    d  tNote, there is no   x in that expression except on the limits. So:      lim          x      →      ∞        ln  ⁡  (  x  )  =      ∫          1        ∞        1    t    d  tThe indefinite integral:This expression:      lim          x      →      ∞        ∫      1    x    d  xDoesn&#039;t actually make sense, because   ∫      1    x    d  x is not well-defined, and, while it is common to refer it as a function of   x,   x is not a free variable.A simple example, with no infinities:      lim          x      →      1        ∫  x    d  x  =      lim          x      →      1                  x      2        2    =      1    2  But:  ∫      (          lim              x        →        1              x    )      d  x  =  ∫  d  x  =  xYou simply can&#039;t swap integrals and limits that way.

Karsyn Beltran · Accepted Answer

It appears you are saying that if      lim          x      →      ∞        f  (  x  )  =  0    ⟹        lim          x      →      ∞            ∫    0    x    f  (  x  )        d    x   convergesin your instance,   f  (  x  )  =      1    x  Although this would make analysis of integrals much simpler, it is not true. There are many functions that vanish at infinity whose integral diverges (      1    x   being a good example).

Proof ' that ln(x) converges Where is the flaw in the following 'proof '? lim_(x ->oo)[(d)/dx{ln(x)}]=lim_(x -> oo}[(1)/(x)]=0 =>lim_(x -> oo)[ln(x)]=text(constant) in bbb(R), therefore ln(x) converges to some real number.

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