Ask a question

Limit of the following:

LIM x→1+       ((1/LN(x)) - (1/x2-1))

2 Answers by Expert Tutors

Tutors, sign in to answer this question.
Andre W. | Friendly tutor for ALL math and physics coursesFriendly tutor for ALL math and physics ...
5.0 5.0 (3 lesson ratings) (3)
Check 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))
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 (638 lesson ratings) (638)
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