A limit problem

  • Thread starter Buri
  • Start date
  • #1
273
0
A limit problem!!

I was trying to calculate the following limit:

lim {x-> infinity} (x² + 2x)^(1/2) - x

I manipulate f(x) in such a way:

f(x) = (x² + 2x)^(1/2) - x
f(x) = [x²(1 + 2/x)]^(1/2) - x
f(x) = |x|(1 + 2/x)^(1/2) - x

As x goes to infinity |x| = x. Therefore,

f(x) = x( [1 + 2/x]^(1/2) - 1 ]

After taking the limit I'm getting infinity times 0 which would mean the limit is equal to zero (I think). However, this isn't the right answer and the way I was supposed to do it was multiply f(x) by its "conjugate" to get the limit as equal to 1. But, I've tried figuring out why I can't do the above, but I just don't see it. So I'd appreciate the help.
 

Answers and Replies

  • #2
212
0


...I'm getting infinity times 0 which would mean the limit is equal to zero (I think).

Here is your mistake. You must remember that infinity is not a number.

in your function f(x) = x( [1 + 2/x]^(1/2) - 1 ] you have, as you point out, one portion going to infinity and the other going to zero. Think of it this way:

How, in a limit sense, can you represent infinity? Easy: lim {x-> infinity} x

How, in a limit sense, can you represent zero? Easy: lim {x-> infinity} 1/x

What happens when you combine the two?

lim {x-> infinity} x* 1/x

which is the same as:

lim {x-> infinity} 1

which is 1.

Now, the above is not how you would prove lim {x-> infinity} (x² + 2x)^(1/2) - x=1.

Have you tried multiplying the function by its conjugate (do you know what that means)?
 
Last edited:
  • #3
2,981
5


Multiply and divide by [itex]\sqrt{x^{2} + 2 x} + x[/itex] and use the rule [itex](a - b)(a + b) = a^{2} - b^{2}[/itex] in the denominator.
 
  • #4
CompuChip
Science Advisor
Homework Helper
4,306
47


After taking the limit I'm getting infinity times 0 which would mean the limit is equal to zero

There is your problem. "Infinity times zero" is an undefined number, precisely because you would be able to get it equal to anything you'd like. For example, consider the limit
[tex]\lim_{x \to \infty} \left( x \cdot \frac{a}{x} \right)[/tex]
which would be equal to 0 by your reasoning (x goes to infinity while a/x goes to 0 for any number a, including a = 0)

Of course, first simplifying, you see that it is
[tex]\lim_{x \to \infty} \frac{a x}{x} = \lim_{x \to \infty} a = a[/tex]

Moral to the story: Whenever you encounter undefined forms in limit operations (like 0/0, or 0 . infinity) you need to exercise more care.
 
  • #5
Mute
Homework Helper
1,388
10


It's already been explained why infinity times zero isn't necessarily zero. The usual way to deal with indeterminate forms is to use L'Hopital's rule. If

[tex]\lim{x\rightarrow \infty} f(x)g(x) = \infty 0[/tex],

then consider f(x)/(1/g(x)). The limit of this is [/itex]\infty/\infty[/itex], so by L'Hopital's rule

[tex]\lim_{x\rightarrow \infty} f(x)g(x) = \lim_{x\rightarrow \infty} \frac{f(x)}{1/g(x)} = \lim_{x\rightarrow \infty} \frac{f'(x)}{(1/g(x))'}[/tex].

There are other ways to evaluate the limit. You were asked to use a conjugate method, which Dickfore gave. Another method is to consider that for x positive

f(x) = x( [1 + 2/x]^(1/2) - 1 ]

is exact. Since 1/x is small, though, you can expand the square root function [itex](1 + 2/x)^{1/2}[/itex] in a taylor series about 1/x = 0. Try that too. (It is easier than L'Hopital's rule).
 
  • #6
273
0


Thanks a lot guys! I had actually done it by multiplying it by its conjugate, but I didn't see why I couldn't do it the way I did it above. I hadn't noticed that lim x * a/x shows my reasoning is incorrect. Thanks a lot for the help!
 
  • #7
273
0


There is your problem. "Infinity times zero" is an undefined number, precisely because you would be able to get it equal to anything you'd like. For example, consider the limit
[tex]\lim_{x \to \infty} \left( x \cdot \frac{a}{x} \right)[/tex]

Just another question. Why is it that I can't break the above limit as follows:

lim x * lim a/x

Is it because lim x doesn't exist? Does the limit property for the product of two functions only apply when the individual limits exist?
 
  • #8
CompuChip
Science Advisor
Homework Helper
4,306
47


Just another question. Why is it that I can't break the above limit as follows:

lim x * lim a/x
Precisely for the reason I show in my post: it would give an undeterminate form (0 x infinity).

Is it because lim x doesn't exist? Does the limit property for the product of two functions only apply when the individual limits exist?

Yep, it is true that if
[tex]\lim_{x \to a} f(x) = L_f[/tex]
and
[tex]\lim_{x \to a} g(x) = L_g[/tex]
where Lf and Lg are numbers (even zero is allowed, but - again - infinity is not a number) then
[tex]\lim_{x \to a} f(x) \cdot g(x) = L_f \cdot L_g[/tex]
and
[tex]\lim_{x \to a} f(x) / g(x) = L_f / L_g[/tex]
(as long as Lg is not zero).
 
  • #9
273
0


Okay, thanks for your help! :)
 
  • #10
2,981
5


Just another question. Why is it that I can't break the above limit as follows:

lim x * lim a/x

Is it because lim x doesn't exist? Does the limit property for the product of two functions only apply when the individual limits exist?

Yes.
 

Related Threads on A limit problem

  • Last Post
Replies
4
Views
673
  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
13
Views
2K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
2
Replies
30
Views
4K
Top