limits -urgent?

Question

find the value of lim x->inf (( ln x)^2)/ x
(a) - inf (b) -1 (c) 0 (d) e (e) none

Answer 1

Answer is:

(c) 0


Proof:


Try to use simple substitution, but in this case it gets difficult - so go for the L'Hospital Rule. Differentiate the numerator & denominator:

lim(x->inf) [ {ln(x)^2} / x ]
= lim (x->inf) [ 2*ln(x)*(1/x) / 1 ]
= lim (x->inf) [ 2*ln(x) / x ]

Again this is not sufficient for us, so apply the rule again:

= lim (x->inf) [ 2*(1/x) / 1 ]
= lim (x->inf) [2 / x]

Apply the limits now,

= [2 / inf] = 0


Hope this helps

Answer 2

A easy possibility is to use l'Hopital Rule. If you differentiate the numerator and the denomonator, you get (2 lnx (1/x))/1 = 2 lnx/x. We still have a limit of the oo/oo kind. If we diffreentiate again, we get (2/x)/1 = 2/x, which clearly goes to 0 when x-> oo.

But if you don't want to use it, put ln x = t, so that x = e^t. Since ln increases monotonically to oo, t goes to infinity when x goes. So, we get lim ( t-> oo) t^2/e^t. We know the exponencial function is of higher order than any polynomila function, so that lim ( t-> oo) t^2/e^t = 0.

The answer is (c)

Answer 3

Since the limit of the numerator will rise slower than the denominator the limit of the function will approach 0. This is because when you have smaller numbers in numerator and larger numbers in denominator the fraction created will get smaller and smaller.

Answer 4

Take derivative of numerator and denominator to get:
2lnx/x This still gives infinity/infinity so do it again to get :
2/x This gives 0 as the limit

Answer 5

2logx.2x/1
4xlogx/1
answer none