How in gods name can: lim x -> infinity x * ln( (x-1) / (x+1) ) = -2 be true?

Question

I cannot see by any means how this limit is true. How can it not be infinity??

Answer 1

actually, the limit is -2, as stated by you.........why is it - 2, you ask ???

1st..... graphic calculator shows limit is - 2

2nd.....the problem is Temporarily Indeterminate [ TI ].... the natural log ratio approaches 1... and ln 1 = 0, so your limit becomes infinity * 0 ......which is one of the 7 indeterminate forms...... [ [ by the way, ln [ ( x - 1 ) / ( x + 1) ] = ln [ 1 - 2 / ( x + 1 ) ], which approaches ln 1 = 0, as x ---> infinity... ]]


you need to rewrite as lim x -->inf { [ ln( ( x - 1 )/( x + 1) ) ] / ( 1 / x ) } , then take the limit...

you get 0 / 0 , TI, and use L'hopital.....

lim x--->inf { [ ( ( x + 1 ) / ( x - 1 ) ) * ( 2 / ( x + 1 )^ 2 ) ] / ( - 1 / x^2) } , or
lim x--->inf [ ( -2 x^2 ) / ( x^2 - 1 ) ] ... after using l'hopital 2 more times, [ limit of - 4x / 2x ], then limit of [ - 4 / 2 ],...both as x ---> infinity... . you get.... -2 ... as an answer ...
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deriv of ln [ ( x - 1 ) / ( x + 1 ] =
[ ( x + 1 ) / ( x - 1 ) ] * [ [ ( x + 1 ) - ( x - 1 ) ] / ( x + 1 )^2 ] =
2 / { ( x - 1 )( x + 1) } = 2 / ( x^2 - 1 )

deriv of 1 / x .... - 1 / x^2
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don't forget to choose a BEST answer, since we spent time helping you....

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edit..... I had one typo on earlier post.. ..... but I have corrected it .... my solution is correct , as it was before the typo ...... a previous answer posted above mine had several typos and has been removed ....

Answer 2

Lots of typos above ...

Rewrite as a fraction with numerator: ln ((x-1)/(x+1)) and denominator: 1/x.

Trot out L'Hopital's rule, differentiating both the numerator and denominator.

The numerator is ln(x-1) - ln(x+1), which has derivative 1/(x-1) - 1/(x+1) = 2/(x^2-1). The denominator has derivative -1/x^2. So their quotient of the derivatives = - x^2 * 2/(x^2-1) = -2/(1 - 1/x^2).

And yes, that has limit -2. Hence so does the original function.

Answer 3

Use L'medical institution Rule lim f(x)/g(x) = f '(x)/g'(x) ; while f(x)/g(x) =infinity/infinity or 0/0 so lim ln(x)/x = lim (a million/x)/a million lim ln(x)/x = lim a million/x lim ln(x)/x = lim a million/infinity lim ln(x)/x = 0

Answer 4

I don't think it IS true: it looks to me as though it would be 0. As x gets big, the argument of the log goes toward 1. so the log approaches 0. This is bait for l'Hopital's rule, try it and see what happens. It's late at night here, so I'm not going to try it myself.