How would you solve the following problem with limits ...

My Answer of the Question of George Stoica Anna Valkova Tomova [email protected]

George Stoica 43.25 Saint John

How would you solve the following problem with limits of functions? Please see the attachment

My Answer Your problem can be summarized in the case when the function has limitless values when the independent variable tends to infinity. We consider the functions,

f '( x ) ln  f ( x )      ln  f ( x)   '  lim f ( x)  lim xf '( x)  0 . provided that: lim     lim x  x  x  f ( x ) 1 ln x  ln x  '    x x Example. Let us consider the function: f  x   ln  x  , x  e  f '( x ) 

lim

x 

ln  ln  x   ln x



ln  ln  x        lim    x  ln x  '



1 . We have: x

1 x ' 1 ln( x )  lim  lim 0 x  x  1 ln  x  x

1 x xf '( x ) x  1  0 . We can verify that:  f ( x ) ln  x  ln  x  x  ln(1  g ( x )  ln x  1   ln  x  xg ( x )  ln  x(1  g ( x )  ln x  ln(1  g ( x )) ln x   0 lim  lim  lim  lim x  x  x  x  ln x ln x ln x ln x because:

 lim g ( x )  1  1  g ( x )  0, x  ,  ln 1  g ( x )  , x 

x  xg ( x )  x(1  g ( x ))  , x   We denote these differentiable functions with the signature:

f ( x )  eo ( f ( x ),ln( x )) , o( f ( x ),ln( x ))  0  ln f ( x )    x, f ( x )   0 , x  ;   x, f ( x )   0 ln( x )





ln f ( x )    x, f ( x )  ln  x   ln x   x , f ( x )   f ( x )  x   x , f ( x )   o( f ( x ),ln( x ))    x, f ( x )  ln  x  Then we have too:

 lim g ( x )  1  1  g ( x )  0, x  ,  ln 1  g ( x )  , x 

x  xg ( x )  x(1  g ( x ))  , x   ln( f ( x  xg ( x )))  ln( f ( x(1  g ( x )))  o(ln( x(1  g ( x ))  o(ln( x )  ln(1  g ( x )) ln( f  x  xg ( x ) )    x  xg ( x ), f  x  xg ( x )    0 , x  , ln  x  xg ( x )  f  x  xg ( x )   eo (ln( x )ln(1 g ( x )) We will prove the same fact: if lim f ( x)  , f '( x), f ( x)  0 and x 

 ln  f ( x )  '

lim

ln  f ( x )         lim ln x    x

lim

f  x  xg ( x )   1,  lim g ( x )  1 x f ( x)

x 

x 

 ln x  '

f '( x ) xf '( x ) f ( x)  lim  lim  0 , then x  x  f ( x ) 1 x

Proof We have: lim

x 

ln

f  x  xg ( x )  f  x  xg ( x )   1  lim ln 0 x  f ( x) f ( x)

f  x  xg ( x )   ln  f  x  xg ( x )    ln f ( x )  f ( x)

ln  f  x  xg ( x )   ln  x  xg ( x )  ln  x  xg ( x ) 



ln f ( x ) ln( x )  ln( x )

  x  xg ( x ), f  x  xg ( x )   ln  x  xg ( x )     x, f ( x )  ln  x  We will consider 3 cases:

1.

lim   x, f ( x )     x  xg ( x ), f ( x  xg ( x ))    0  x 

  x, f ( x )     x  xg ( x ), f ( x  xg ( x ))  , x  

2.   x, f ( x)     x  xg ( x), f ( x  xg ( x))  3.   x, f ( x)     x  xg ( x), f ( x  xg ( x))  . 1.

If lim   x, f ( x )     x  xg ( x ), f ( x  xg ( x ))    0  x 

  x, f ( x )     x  xg ( x ), f ( x  xg ( x ))  , x   we have

  x  xg ( x ), f  x  xg ( x )   ln  x  xg ( x )     x, f ( x )  ln  x     x  xg ( x ), f  x  xg ( x )    ln  x  xg ( x )   ln  x    x  xg ( x )  x   x  xg ( x ), f  x  xg ( x )   ln 1  g ( x )   0, x  

  x  xg ( x ), f  x  xg ( x )   ln

2.

  x, f ( x)     x  xg ( x), f ( x  xg ( x))  we have:

  x  xg ( x ), f ( x  xg ( x ))  ln  x  xg ( x )     x, f ( x )  ln  x  



ln  x  xg ( x )  ln x

  x  xg ( x ), f ( x  xg ( x )) 

  x  xg ( x ), f ( x  xg ( x ))     x , f ( x ) 

1  g ( x ) 

  ln  x

1  g ( x ) 

  x , f ( x )



 x  xg ( x )    ln

 x  xg ( x ), f ( x  xg ( x ))    x , f ( x )

x

  x  xg ( x ), f ( x  xg ( x )) 

 0,

  x  xg ( x ), f ( x  xg ( x )) 

 0, x  , x   x , f ( x )   x  xg ( x ), f ( x  xg ( x ))    x, f ( x )     x  xg ( x ), f ( x  xg ( x ))  ln

3.

if   x, f ( x)     x  xg ( x), f ( x  xg ( x))  we have:

lim

x 

ln 

f  x  xg ( x )  f ( x)  1  lim ln 0 x  f ( x) f  x  xg ( x ) 

f ( x)   ln  f  x  xg ( x )    ln f ( x )  f  x  xg ( x ) 

ln  f  x  xg ( x )   ln  x  xg ( x )  ln  x  xg ( x ) 



ln f ( x ) ln( x )  ln( x )

  x  xg ( x ), f  x  xg ( x )   ln  x  xg ( x )     x, f ( x )  ln  x  But we have:



  x  xg ( x ), f ( x  xg ( x ))  ln  x  xg ( x )     x, f ( x )  ln  x  



 ln  x  xg ( x )  ln x

  x  xg ( x ), f ( x  xg ( x )) 

   x  xg ( x ), f ( x  xg ( x ))     x , f ( x ) 

  ln  x

1  g ( x) 

  x , f ( x )

  ln  x  xg ( x)

   x  xg ( x ), f ( x  xg ( x )) 

1

ln

  x  xg ( x ), f ( x  xg ( x )) 

  x  xg ( x ), f ( x  xg ( x ))     x , f ( x ) 

x 1  g ( x )    x, f ( x )     x  xg ( x ), f ( x  xg ( x )) 

  x , f ( x )

x

  x  xg ( x ), f ( x  xg ( x )) 



 0,

 0, x  ,

The proof is complete. Remark. We can explorer more examples, considering the functions:

  f ( x )  ln(ln(...( x )...)  ln f ( x)  o  ln(ln(...( x)...)  ,   n n   e ... ln f ( x )    x, f ( x )   0, x  , x  e e  x    , ln(ln(...( x )...) lim f ( x )  , lim x 

x 

xf '  x  0 f ( x)

which satisfy this condition, that lim x 

f  x  g ( x) x   1 . We can define for any f ( x)

function which satisfies the condition the same signature:

lim x 

ln f ( x ) ln f ( x ) 0    x, f ( x )   0, x    ln(ln(...( x )...)) ln(ln(...( x )...)) n

ln( f ( x ))  o(ln(ln(...( x )...))), x  ,  ln f ( x)    x, f ( x)  ln(ln(...( x )...))  f ( x )  eo (ln(ln(...( x )...))  e

  x , f ( x ) (ln(ln(...( x )...) )) n

e

   ln(ln(...( x )...)  x , f ( x )  )    n 1  

 ln(...( x )...) n 1

and resolve the same problem.

  x , f ( x )