^{+ }((

^{1/}

_{LN(x)})

_{ }- (

^{1}/

_{x2-1)})

LIM x→1^{+ }((^{1/}_{LN(x)})_{ }- (^{1}/_{x2-1)})

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Marked as Best Answer

The limit is of the indefinite form ∞-∞, so you will need L'Hospital's Rule. It says that for indefinite forms,

lim (f(x)/g(x)) = lim (f'(x)/g'(x)).

To apply the rule, you need to find the common denominator:

1/ln(x) - 1/(x²-1) = ((x²-1) - ln(x))/(ln(x) (x²-1))

Then

lim_{x→1+} ((x²-1) - ln(x))/(ln(x) (x²-1))

= lim_{x→1+} ((2x-1/x)/((x²-1)/x+ln(x) (2x))

= lim_{x→1+} ((2x²-1)/(x²-1+2x²ln(x))

This limit is of the form 1^{+}/0^{+}, so it equals +∞.

Richard P. | Fairfax County Tutor for HS Math and ScienceFairfax County Tutor for HS Math and Sci...

A related, more interesting, limit is:

lim_{x->1 }[ 1/ln(x) - 2/(x^{2} -1) ]

L'Hospital's rule will work for this one , but more insight can be had by making the substitution z = x -1

lim_{z->0 }1/ln(1+z) - 2/(2 z + z^{2}) to order z^{2} ln(1+z) = z - z^{2}/2 so we have

lim_{z->0 }1/(z -z^{2}/2) - 1/(z + z^{2}/2) On subtracting these fractions we get to order z^{2}

lim_{z->0 }z^{2} /z^{2} = 1

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