# Solving infinity limit, something weird.

1. Apr 1, 2007

### wk1989

1. The problem statement, all variables and given/known data
This question asks you to evaluate the limit of the following function

lim sqrt(9x^2-3x) -3x
x-> infinity

I did it using 2 methods:
1)

lim sqrt(9x^2-3x) -3x
x-> infinity

= lim x(sqrt(9-3/x)) -3x
x-> infinity

= lim 3x - 3x
x -> infinity

= 0

Second Method:
I multiplied by the conjugate radical/conjugate radical and after simplification, I get

lim -3x/6x
x -> infinity

= 1/2

Why is it that I'm getting different answers? Which solution is the right one? Also, how do I type up mathematical formulas? Can someone point me to a LaTex tutorial? Thanks!

2. Apr 1, 2007

### Mathgician

when you multiplied the conjugate, did you also multiply the denominator?

3. Apr 1, 2007

### Spiro09

your first attempt is wrong because your trying to solve the limit of only one part of your equation without the other.
You can't solve the lim (x-> infinity) of sqrt(9-3x) without solving the rest.

Your second answer is right except for the sign, it should be -1/2.

Last edited: Apr 1, 2007
4. Apr 1, 2007

### wk1989

Alright thanks to both of you, I have another question though. If I evaluate both parts in an equation and end up with

3x - 3x
Where x is approaching infinity, would it be right to say that the limit is 0? I remember my teacher saying that you cannot perform arithmetic operations involving infinity, then how come you can divide out -3x/6x? I'm new to calculus so please help lol.

5. Apr 2, 2007

### Mathgician

if it is 3x- 3x, then yes because 3[ lim x(1-1)]=0

Last edited: Apr 2, 2007
6. Apr 2, 2007

### Mathgician

my bad, what I had on there earlier was a mistake, sorry. Infinity multiplied by zero should equal zero right? So it should approach zero, I think. Can anybody explain clearly the properties of infinity for limits and its relationship with the concept of zero? Hey wk1989, you are asking very intelligent questions, and I felt the same frustration when I took Calc I. They slap me with these concepts and I was to take it in without much clear explanations that I could imagine. If you are going to take Calc. II, when you study series, this kind of stuff will be explained further, so don't worry.

Last edited: Apr 2, 2007
7. Apr 2, 2007

### sutupidmath

noooooo, infinity multiplied by zero is undefined

8. Apr 2, 2007

### Mathgician

why? Can you explain the reasoning behind it?

9. Apr 2, 2007

### sutupidmath

i guess what u really meant is this, couse u forgot a pair of brackets here
this
lim sqrt(9x^2-3x) -3x
x-> infinity
and this

lim (sqrt(9x^2-3x) -3x), x->infinity

are not the same.
however the answer like the others stated is -(1/2)

10. Apr 2, 2007

### cristo

Staff Emeritus
Infinity is not a number (i.e. it is not a member of the real, complex, rational, etc.. numbers) and so multiplication by infinity is undefined.

11. Apr 2, 2007

### cristo

Staff Emeritus
That's a little pedantic isn't it?

12. Apr 2, 2007

### sutupidmath

you are saying that

0*infinity =0, then if this would be right then this also wuld be right

0/0=infinity, if we devide by zero your "equation". but we come to a contradiction here couse we know that 0/0 is undefined. or it also would be right according to what you are saying to conclude that

(0/0)*infinity=0/0, which again is undefined.

somebody else migh give a more effective and detailed reasoning i guess.

13. Apr 2, 2007

### sutupidmath

no i think it's about writting it correctly.

14. Apr 2, 2007

### cristo

Staff Emeritus
In all fairness though, he wrote this $$\lim_{x\rightarrow\infty}\sqrt{9x^2-3x}-3x$$, and you wrote this $$\lim(\sqrt{9x^2-3x}-3x), x\rightarrow\infty$$.

The former is a more standard way of writing it!

15. Apr 2, 2007

### Mathgician

Yeah, that is logical mathematically

16. Apr 2, 2007

### sutupidmath

i am not talking about what is standard and not, but you have to understand that these two expressions are not equivalent, they represent different things. At the former 3x is out of limit when x-> infinity, so it means that 3x at this case is absolutely not affected when x-> infinity, but at the latter one when 3x is inside the breackets it means that we should also consider 3x when we evaluate the limit of the whole expression when x-> infinity.

you cristo are saying this in a more general form

lim f(x)-g(x),x->infinity is equal to lim[f(x)-g(x)],x->infinity, which is absurd. because the latter one using the features of additon and subtraction of limits we get

lim[f(x)-g(x)],x->infinity=lim f(x)-lim g(x),x->infinity, so we get something which is not the same as lim f(x)-g(x),x->infinity , as you are claiming to be. Once more these two limits are not the same.

can someone else coment on this???

Last edited: Apr 2, 2007
17. Apr 2, 2007

### HallsofIvy

Staff Emeritus
Unfortunately, you made this ambiguous by also moving the "$x\rightarrow \infty$"!

I understand that your point is that
$$\lim_{x\rightarrow\infty}(\sqrt{9x^2-3x})- 3x$$
is not the same as
$$\lim_{x\rightarrow\infty}(\sqrt{9x^2-3x}- 3x)$$
But I am not sure that I agree that one must interpret
$$\lim_{x\rightarrow\infty}\sqrt{9x^2-3x}- 3x$$
as the former rather than the latter.

18. Apr 2, 2007

### sutupidmath

I think in order to avoid any misinterpretation or ambiguity, we should write it as i suggested. The way cristo said can also be written like this

lim sqrt(9x^2-3x)-3x=-3x+lim sqrt(9x^2-3x), when x->infinity in both cases. however if we write it this way lim [sqrt(9x^2-3x)-3x] is not equal to -3x+lim sqrt(9x^2-3x), when x->infinity, in both cases again. I am sorry for the symbols.

so this is my poing

Last edited: Apr 2, 2007
19. Apr 2, 2007

### HallsofIvy

Staff Emeritus
Oh, I agree with that! The more parentheses the better!

20. Apr 2, 2007

### cristo

Staff Emeritus
Ok, standard was the wrong word to use, but I don't "have to understand they are not equivalent!" Given this limit $$\lim_{x\rightarrow\infty}\sqrt{9x^2-3x}-3x$$, I (for one) would automatically interpret this as $$\lim_{x\rightarrow\infty}(\sqrt{9x^2-3x}-3x)$$. If one wanted to take the limit of only the first term, then one would put parentheses around only the first term. That was my point, and that's why I said it was a little pedantic to correct something that didn't need correcting!