L'Hospitals Rule: find limit, as x->infty, of ln(x)/sqrt(x)

I just had a question of a test that I missed. How do you do it?

lim lnx/(square root of x)
x->oo

The answer I got was oo which is wrong. I think the answer might be 0 but I'm not sure. How do you do this problem?

Re: Help with L'Hospitals Rule

lim x-->oo lnx/x^1/2= oo/oo undefined
lim x-->oo [1/x] /{1/2 x^(-1/2)]
lim x-->oo 2/{x[x^-1/2]}
lim x-->oo 2/x^1/2 = 2/oo
lim x-->0 lnx/x^1/2 =0 answer

Arthur

Re: Help with L'Hospitals Rule

thatguy47 said:

I just had a question of a test that I missed. How do you do it?

lim lnx/(square root of x)
x->oo

The answer I got was oo which is wrong. I think the answer might be 0 but I'm not sure. How do you do this problem?

Click to expand...

There are two well-known (that means: Well, I know them.) limit properties: (for x -> infinity)

1. The 'e^x grows fast' rule:

lim e^x/x^n = infinity, no matter how big n is.

2. The "ln x grows slow" rule: (Sorry -- slowLY)

lim ln x/x^n = 0, no matter how SMALL n is. (for n > 0)

OR

lim ln x/x^(1/k) = 0 for any (large) k.

Use l'Hospital's rule for either of them.