Sat 15 Oct, 2011 12:26 pm
1. numerator is -infinity, denominator is 0 (essentially -infinity * infinity)
2. y^2 = 2x - x^2 + 9
differentiate both sides to get:
d/dy = 2y
d/dx = 2-2x
dy/dx = (2-2x )/ 2y
lim as x->0 f(x)/g(x)=lim as x>0 f'(x)/g'(x)
d[ln(x)]/dx=1/x and dx/dx=1
lim x->0 ln(x)/x=lim x->0 1/x=inf
L'Hopital's rule gives the wrong answer (wrong sign), and is not required. You don't need LH for (-infinity)/0.
l'Hospital's rule requires the existence of f'(x) and g'(x) respectively, but when x tends to +0, the limit of lnx doesn't exist.
l'Hospital's rule requires existence of f'(x) and g'(x) respectively. but, when x tends to +0, the limit of lnx doesn't exist, so the prerequisite to exert l'Hospital's rule doesn't meet, so it can't be used to solve the problem.
''markr'' solved the problems in right way and ''raprap'' 's use of L'Hospital's law was wrong , but the reason why the law can not be used here is not exactly that presented by ''witswang'' , the L'Hospitals actually is;
when lim as x-->a f(x) & lim as x-->a g(x) are either both 0 or, both infinity
(or, - infinity)
then lim as x-->a f(x)/g(x) = lim as x-->0 f ' (x) / g ' (x) .
The reason why L'Hospital's law can not be used in the given problem is due to the fact that lim as x-->0+ lnx = - infinity but lim as x-->0+ x = 0
so both are not same hence condition of L'Hospital's law is not satisfied .