1. L'Hospital Rule Indeterminate INF-INF

I was just wanting to know if someone could tell me what to do in this circumstance. I do know that I have to get a common denominator, on both sides of the problem, but what if there are no fractions in the problem; how would I take the problem and turn it into a fraction? Thanks for all of the help!

2. Originally Posted by qbkr21
I was just wanting to know if someone could tell me what to do in this circumstance. I do know that I have to get a common denominator, on both sides of the problem, but what if there are no fractions in the problem; how would I take the problem and turn it into a fraction? Thanks for all of the help!
Where's the problem?

3. Re:

I am just thinking in general, but for instance

The limit as x approaches infinity of (xe^(1/2)-x)

4. Originally Posted by qbkr21
I am just thinking in general, but for instance

The limit as x approaches infinity of (xe^(1/2)-x)
you dont need L'Hospital's rule for this one

xe^(1/2)-x=x(e^(1/2)-1)->infty

if you try lim_x->infty (xe^(1/x)-x)

=lim_x->infty x(e^(1/x)-1)=lim_x->infty (e^(1/x)-1)/(1/x)

now use L'Hospital's rule

5. Originally Posted by qbkr21
I am just thinking in general, but for instance

The limit as x approaches infinity of (xe^(1/2)-x)
that's a bad example actually, we wouldn't use l'hopital's for that, it obviously goes to infinity

lim{x-->oo} xe^(1/2) - x
= lim{x-->oo} x(e^(1/2) - 1) ..........(e^(1/2) - 1) is a constant we can put it in front
= (e^(1/2) - 1) lim{x-->oo} x
= (e^(1/2) - 1)*infinity
= infinity

6. Re:

Could you tell me in words what steps you went through? I appreciate the help!

7. Re:

Can you think of a better example.

I learned that 0*INF you put one term on the bottom and INF-INF you take a common denominator on both sides... I understand 0*INF but not INF-INF...

8. Originally Posted by qbkr21
Could you tell me in words what steps you went through? I appreciate the help!
so this is what frenzy did:

if you try lim_x->infty (xe^(1/x)-x)

=lim_x->infty x(e^(1/x)-1)=lim_x->infty (e^(1/x)-1)/(1/x)

now use L'Hospital's rule

what are the steps?

note that the way i worked out the first example would not work here, we would get 0*infinity, which is what you asked about. so here's how frenzy turned it into a quotient.

lim_x->infty (xe^(1/x)-x) ..........we see this is not a nice function, try to simplify a bit.
= lim_x->infty x(e^(1/x)-1)

ok, now we have 0*infinity if we take the limit, to turn this into a quotient, we have to find some way to put one of the numbers in the bottom. changing it into a fraction where (e^(1/x)-1) is in the bottom is too much trouble, so let's put the x in the bottom, but how? so after looking really hard and thinking about this for days we realize x = 1/(1/x), so we replace it with that, so we have

= lim_x->infty (1/(1/x))(e^(1/x)-1) .......putting this in a fraction we get
= lim_x->infty (e^(1/x)-1)/(1/x) ........now this goes to 0/0 which is a condition for using l'hopital's.

so it might take a while to see how frenzy put x in the bottom the first time, but afterwards, you remember it forever. this is a standard trick actually, and it happens often in limits (changing x to 1/(1/x)), so remember it well

9. Originally Posted by qbkr21
Can you think of a better example.

I learned that 0*INF you put one term on the bottom and INF-INF you take a common denominator on both sides... I understand 0*INF but not INF-INF...
it's kind of hard to come up with an example off the top of my head for INF - INF. But we can get it by twisting the problem we already have (boy, we're getting a lot of use out of this!)

consider lim{x-->oo} xe^x - x, obviously this goes to INF - INF. the thing is, this is not an example to show off the method you asked about, about combining denominators. we would do this problem exactly the same way we did the last. that is, factor out the x, change it to 1/(1/x) and use l'hopital's

10. Re:

Thanks Jhevon! So let me get this straight. If I am able to factor out an X then I can create a quotient by sticking it in as the denominator, but really it is (1/x)? What if it had a constant in front of it would I leave it alone and just take the X down to the bottom? Sorry I have been off and on tonight I have a huge History Paper due on Mon.

11. Originally Posted by qbkr21
Thanks Jhevon! So let me get this straight. If I am able to factor out an X then I can create a quotient by sticking it in as the denominator, but really it is (1/x)? What if it had a constant in front of it would I leave it alone and just take the X down to the bottom? Sorry I have been off and on tonight I have a huge History Paper due on Mon.
well, it is 1/(1/x) which is x, written differently, when we combine the function it becomes f(x)/(1/x)

whenever constants are in front you don't really have to worry about them, you can just move them in front of the limit sign and continue as normal, and multiply out when done

12. Re:

Ohh... I thought this problem was a INF-INF problem. I guess I am just trying to figure out how with the subtraction in this problem it yields a 0*INF case...

13. Re:

YES THANKS JHEVON!!!I figured it out. I didn't factor when I initially looked at this problem...

Therefore I thought (xe^(1/x)-x) => ((INF)e^(1/(INF))-(INF)) = INF-INF

Thus I was look at the wrong case.

I think that they key to solving these is factoring and good simplification before you do anything with the problem. Thanks Again!

14. Just in case, here's a beauty limit if you wanna practice

15. Originally Posted by qbkr21
YES THANKS JHEVON!!!I figured it out. I didn't factor when I initially looked at this problem...

Therefore I thought (xe^(1/x)-x) => ((INF)e^(1/(INF))-(INF)) = INF-INF

Thus I was look at the wrong case.

I think that they key to solving these is factoring and good simplification before you do anything with the problem. Thanks Again!
well, you are actually correct here. e^(1/INF) --> e^0 = 1

so ((INF)e^(1/(INF))-(INF)) = ((INF)(1)-(INF)) = INF-INF

But ok, now you know how to do it

for the problem Krizalid posted (I can understand the spanish in there by the way --but it's common sense still so i wont be too happy about it ) i would evaluate it using logarithms...just a hint to get you started

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