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2 Answers

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 +∞.
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