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# Limit of the following:

LIM x→1+       ((1/LN(x)) - (1/x2-1))

### 2 Answers by Expert Tutors

Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
1
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

limx→1+ ((x²-1) - ln(x))/(ln(x) (x²-1))
= limx→1+ ((2x-1/x)/((x²-1)/x+ln(x) (2x))
= limx→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...
4.9 4.9 (617 lesson ratings) (617)
-1
A related, more interesting, limit is:

limx->1 [ 1/ln(x) - 2/(x2 -1) ]

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

limz->0 1/ln(1+z) - 2/(2 z + z2)        to order z2   ln(1+z) = z - z2/2    so we have

limz->0 1/(z -z2/2) - 1/(z + z2/2)      On subtracting these fractions we get to order z2

limz->0 z2 /z2       =   1