Limit of the following:

Question

LIM x→1⁺       ((^1/_LN(x)) - (¹/_x²-1))

Answer 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

 

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 +∞.

Answer 2

A related, more interesting, limit is:

 

lim_x->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

 

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

 

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

 

lim_z->0 z² /z²       =   1