What is the limit of ln(x)-square root(x) when x approaches +oo ?

If you directly try to take the limits we get the form ∞ - ∞. In order to use L'Hospital's Rule, we will need to express the terms in such a way that we have lim x →∞ [ f(x)/g(x) ] = 0/0 or ∞/∞.

We can do so by manipulating the terms into fraction form.

lim x →∞ [ ln(x) - sqrt(x) ]

= lim x →∞ [ (ln(x) - sqrt(x))( ln(x) + sqrt(x)) / (ln x + sqrt(x)) ]

= lim x →∞ [ (2 ln(x) ) / ( ln(x) + sqrt(x)) ] - lim x →∞ [ (x) / ( ln(x) + sqrt(x)) ]

This is now of the form ∞/∞ so we may use L'Hospital's Rule.

Getting the derivatives of the numerator and denominator we obtain:

= lim x →∞ [ (2/x) / ( (1/x) + (0.5)( 1/sqrt(x)) ) ] - lim x →∞ [ 1 / ( (1/x) + (0.5)( 1/sqrt(x) ) ) ]

= lim x →∞ [2 / (1 + 0.5(sqrt(x)) )] - lim x →∞ [x/(1 + 0.5(sqrt(x)) )]

*See that lim x →∞[ [2 / (1 + 0.5(sqrt(x)) )] = 0 while lim x →∞ [x/(1 + 0.5(sqrt(x)) )] = ∞/∞ so we use L'Hospital's Rule a second time on lim x →∞ [x/(1 + 0.5(sqrt(x)) )].

= 0 - lim x →∞ [1/( 1 + 0.25/(sqrt(x)) )]

= 0 - ∞

= -∞