# Homework Help: L'Hopital! problem help

1. Dec 8, 2009

### IntegrateMe

Estimate the limit using L'Hopital's rule:

1. limit of x is going to infinity of xtan(1/x)

Okay, I have no idea how to do #1, but i know the indetermiate form is infinity*0.

2. limit as x tends to 0+ of (lnx - ln sinx)

For #2 the indeterminate form is infinity-infinity, so i took the derivative of the function and got:

(1/x) - (cosx/sinx) which yields the same indeterminate form of infinity - infinity.

So my question is, should i get both with a common denominator, or should i derive (1/x) - (cosx/sinx) again and plug 0 to see if it works?

Any suggestions? For both 1 or 2?

2. Dec 8, 2009

### Staff: Mentor

Re: L'Hopital!

1. x*tan(1/x) = [tan(1/x)]/[1/x]
If you let u = 1/x, then the above is tan(u)/u which has a known limit as u -->0+.

2. lnx - ln sinx = ln(x/sin x)

lim ln(x/sin x) = ln(lim (x/sin x)), provided that the limit exists.

3. Dec 8, 2009

### IntegrateMe

Re: L'Hopital!

Mark, tan(u)/u is going to be 0/0 which is an indeterminate form?

4. Dec 8, 2009

### Staff: Mentor

Re: L'Hopital!

Do you have a question about what I said?

5. Dec 8, 2009

### IntegrateMe

Re: L'Hopital!

Well, tan(u)/u is 0/0 which is an indeterminate form. Do i take the derivative of tan(u)/u then substitute "u" back into the equation? I'm confused on what you did for #2?

6. Dec 8, 2009

### physicsman2

Re: L'Hopital!

for 1, in my opinion, i don't like u substitution so I just evaluate the limit without substituting.
I would do what Mark44 said and form tan(1/x)/(1/x)
as you should see, you'll get 0/0 for which you can take the derivatives and go from there

7. Dec 8, 2009

### physicsman2

Re: L'Hopital!

for number 2, the properties of a logarithm were applied and since ln is a continuous function, you can take the limit of the inside while keeping ln on the outside for a little

8. Dec 8, 2009

### IntegrateMe

Re: L'Hopital!

Ok so, dy/dx = [sec2(1/x)/x2]/(1/x2)
Then what?

Any help for #2?

9. Dec 8, 2009

### IntegrateMe

Re: L'Hopital!

So, for #2 i can do:

d/dx ln(lim (x/sin x)) = ln (1/cosx) ?
plug in 0 and get ln (1/1) = 0?

10. Dec 8, 2009

### physicsman2

Re: L'Hopital!

for number 2, again do what Mark44 suggested without any substitution(my opinion) if the need arises.
So, ln(x) - ln (sinx) becomes ln(x/sinx)
(remember that property of logarithms)(if not, it would be best to look over those again)

then, you take the limit as x approaches 0 from the right of ln(x/sinx).
since ln is a continuous function on its entire domain(looking at the graph of ln x should answer any doubts), you can first take the limit as x approaches 0+ of x/sinx, then when you get an answer put it in the ln function like so: ln(answer) = true answer ( can't forget about the ln you took out)

11. Dec 8, 2009

### physicsman2

Re: L'Hopital!

are you sure 1/1 = 0? and don't forget about the ln you took out to apply the limit

12. Dec 8, 2009

### IntegrateMe

Re: L'Hopital!

i said the ln(1/1) = 0

so if i were to use the ln(x/sinx) function and find the limit as it tends to 0+, i would just get the ln(0/0); so, the answer is just infinity?

13. Dec 8, 2009

### physicsman2

Re: L'Hopital!

good question
i would say no because you have an indeterminate form, which wouldn't be infinity
you need to resolve the problem of the indeterminate form to evaluate the limit
0/0 is definitely not infinity

14. Dec 8, 2009

### IntegrateMe

Re: L'Hopital!

Oh, i see.
Okay, i think i understand the situation now, thanks!