x->-infinity,lim(x+sqrt(x^2-4x+1))

 

 

i can't for the life of me work out why or how it is possible the limit is 2?!?!

Edit: Sorry, this was wrong. Thought you said +infinity instead of -infinity

My suggestion in limit solving. Divide the stuff inside the limit by x, you will get: 1 + sqrt(1 - 4/x + 1/x^2). Then substituting in x -> -infinity, you will get 2.

Huh? If you divide the limit by x, you're also going to need to multiply it by x to keep equality on both sides. That method works well for fractions, but in this case, I think you need something else.

Huh? If you divide the limit by x, you're also going to need to multiply it by x to keep equality on both sides. That method works well for fractions, but in this case, I think you need something else.

 

 

Aww. you're right XD. I guess I couldn't take the easy way out. Let me think about it ~_~

x->-infinity' date='lim(x+sqrt(x^2-4x+1))

 

 

i can't for the life of me work out why or how it is possible the limit is 2?!?![/quote']So, we wish to calculate the [math]\lim_{x\to - \infty}{x+\sqrt{x^2-4x+1}[/math][math] = \lim_{x\to\infty}{-x+\sqrt{x^2+4x+1}}[/math].

 

We'll employ a standard trick, multiplication by the 'conjugate'.

 

Note that:

 

[math]-x+\sqrt{x^2+4x+1} = (-x+\sqrt{x^2+4x+1}) \cdot[/math][math]\frac{- x - \sqrt{x^2+4x+1}}{ - x - \sqrt{x^2+4x+1}}=(1)[/math]

 

[math] = \frac{x^2 - (x^2+4x+1)}{-x - \sqrt{x^2+4x+1}} = \frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math] = \frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4+1/x}{1+\frac{1}{x} \sqrt{x^2+4x+1}}[/math][math]=\frac{4+1/x}{1+\sqrt{(\frac{1}{x^2})(x^2+4x+1)}}[/math]

 

[math]=\frac{4+1/x}{1+\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}}[/math]

 

Thus,

 

[math]\lim_{x\to-\infty}{x+\sqrt{x^2-4x+1}=\lim_{x\to\infty}{\frac{4+1/x}{1+\sqrt{1+\frac{4}{x}+\frac{1}{x^2}}}}[/math][math]=\frac{4}{1+\sqrt{1+0+0}}=2[/math]

 

 

Alternatively, we could have used L'Hopital's rule after simplifying step [math](1)[/math], but I wasn't sure if you had learned it as of yet.

 

Edit: On a related note, apparently the [math]\LaTeX[/math] renderer here has a length limit. I had to break up quite a few of the above statements into separate [ math] [/ math] chunks.

Ah there you go One of these days I should learn how to use Latex

So' date=' we wish to calculate the [math']\lim_{x\to\infty}{x+\sqrt{x^2-4x+1}[/math].

 

We'll employ a standard trick, multiplication by the 'conjugate'.

 

Note that:

 

[math]x+\sqrt{x^2-4x+1} = (x+\sqrt{x^2-4x+1}) \cdot \frac{x - \sqrt{x^2-4x+1}}{x - \sqrt{x^2-4x+1}}(1)[/math]

 

[math] = \frac{x^2 - (x^2-4x+1)}{x - \sqrt{x^2-4x+1}} = \frac{4x-1}{x - \sqrt{x^2-4x+1}} = \frac{4x-1}{x - \sqrt{x^2-4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4-1/x}{1-\frac{1}{x} \sqrt{x^2-4x+1}}[/math][math]=\frac{4-1/x}{1-\sqrt{(\frac{1}{x^2})(x^2-4x+1)}}[/math]

 

[math]=\frac{4-1/x}{1-\sqrt{1-\frac{4}{x}+\frac{1}{x^2}}}[/math]

 

Thus,

 

[math]\lim_{x\to\infty}{x+\sqrt{x^2-4x+1}=\lim_{x\to\infty}{\frac{4-1/x}{1-\sqrt{1-\frac{4}{x}+\frac{1}{x^2}}}}[/math][math]=\frac{4}{1-\sqrt{1-0+0}}=2[/math]

 

 

Alternatively, we could have used L'Hopital's rule after simplifying step [math](1)[/math], but I wasn't sure if you had learned it as of yet.

 

Edit: On a related note, apparently the [math]\LaTeX[/math] renderer here has a length limit. I had to break up quite a few of the above statements into separate [ math] [/ math] chunks.

 

Dapthar this utterly confused me. Firstly the limit is towards negative infinity, not positive infinity. Second of all, in the denominator you have 1-1=0, which is division by zero error. Not two in the denominator.

 

You seem to have written her problem down incorrectly. Let me give it a try:

 

[math] \lim_{x\to - \infty} [x + \sqrt{x^2-4x+1}] [/math]

 

So you want to know what happens to the value of the function you are taking the limit of, as x becomes increasingly negative.

 

The first thing I would do, is replace x by -x, and take the limit as x goes to positive infinity. So you have this instead...

 

[math] \lim_{x\to \infty} [-x + \sqrt{x^2+4x+1}] [/math]

 

Intuitively, x^2 dominates over 4x. In other words, as x becomes increasingly huge, x^2 becomes enormous in comparison to 4x, so that we can neglect the addition of 4x to x^2, and certainly adding 1 doesn't alter the answer much either. Thus, we should expect the limit above to be equal to the following limit:

 

[math] \lim_{x\to \infty} [-x + \sqrt{x^2}] = \lim_{x\to \infty} [-x + x] = \lim_{x\to \infty} [0] = 0 [/math]

 

Now, of course this is not rigorous, but it's intuitive.

 

And one other thing, the square root of x^2 is the absolute value of x.

Dapthar this utterly confused me. Firstly the limit is towards negative infinity' date=' not positive infinity. Second of all, in the denominator you have 1-1=0, which is division by zero error. Not two in the denominator.

[/quote']You're right Johnny, I'll go back and fix my post.

You're right Johnny, I'll go back and fix my post.

No problem Dapthar, but...

 

Isn't the limit zero?

o.O

 

The problem with the negative infinity can be easily resolved with the t = -1/x substitution, where as x -> negative infinity, t -> 0. If you do this substitution on his third line will give you good results.

 

An alternative method I thought of was to actually factor the insides of the square root before doing all of that, you will find that it turns out to be something like: (x-a)(x+a), where a = 2 (+-) sqrt3. Then from there doing the same, you'll end up with a cleaner process.

o.O

 

The problem with the negative infinity can be easily resolved with the t = -1/x substitution' date=' where as x -> negative infinity, t -> 0. If you do this substitution on his third line will give you good results.

 

An alternative method I thought of was to actually factor the insides of the square root before doing all of that, you will find that it turns out to be something like: (x-a)(x+a), where a = 2 (+-) sqrt3. Then from there doing the same, you'll end up with a cleaner process.[/quote']

 

Yes I thought about that too.

 

(x+a)(x+b) = x^2 +xb+ax+ab

 

ab=1

 

x(a+b)=4x

(a+b) =4

 

b=1/a

 

(a+1/a)=4

 

(a^2+1)/a = 4

 

a^2+1=4a

a^2-4a+1=0

 

Now use quadratic formula. By a theorem due to Abel, there are exactly two roots of the polynomial above. The roots of the polynomial above are given by:

 

[math] \frac{4 + \sqrt{16-4}}{2} [/math]

 

[math] \frac{4 - \sqrt{16-4}}{2} [/math]

 

Now

 

[math] \sqrt{16-4} = \sqrt{12} = \sqrt{3*2*2} = 2 \sqrt{3} [/math]

 

Thus, we can write the two roots in a simpler form:

 

[math] \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}[/math]

 

[math] \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3} [/math]

 

Let us write this:

 

[math] r1 = 2 + \sqrt{3}[/math]

 

[math] r2 = 2 - \sqrt{3} [/math]

 

So if a=r1 then the polynomial statement is true, and if a=r2, then the polynomial statement is true, and there are no other numbers which when instantiated for the variable a... yield a true statement.

Okay, here's my kind of "sketch" proof:

 

[math]x^2 - 4x + 1 = (x-2)^2 - 3[/math]

 

Now let [math]y = x-2[/math], so the limit becomes:

 

[math]\lim_{y\to -\infty} \left( 2 + y + \sqrt{y^2 -3} \right)[/math]

 

Now, [math]2 + y + \sqrt{y^2 -3} = 2 + \frac{3}{y - \sqrt{y^2 - 3}}[/math]

 

The fraction on the right tends to zero, so the limit is 2. I did it Dapthar's way at first, but got unstuck at the end also. What do you guys think? Personally, I think it's a little bit dodgy and I haven't actually proved that the fraction on the right tends to zero, but I think that it's pretty easy to do so (sandwich rule?)

Edit: On a related note, apparently the [math]\LaTeX[/math] renderer here has a length limit. I had to break up quite a few of the above statements into separate [ math] [/ math][/b'] chunks.

 

Indeed. The limit is about 500 pixels to stop the page breaking up and whatnot.

Ok, I fixed my earlier calculation. (See above post.)

Ok, I fixed my earlier calculation. (See above post.[/i'])

 

There are still a few errors Dapthar. Look at the sign of your 4x term. In one of the steps you switch from negative to positive by accident.

There are still a few errors Dapthar. Look at the sign of your 4x term. In one of the steps you switch from negative to positive by accident.

I cancel out the [math]-1[/math] from the numerator and the denominator in this step:

 

[math]\frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4+1/x}{1+\frac{1}{x} \sqrt{x^2+4x+1}}[/math]

 

If that's the error you're referring to.

I cancel out the [math]-1[/math] from the numerator and the denominator in this step:

 

[math]\frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4+1/x}{1+\frac{1}{x} \sqrt{x^2+4x+1}}[/math]

 

If that's the error you're referring to.

 

 

That's not an error. Here i will show you what I'm referring to...

[math] \frac{-4x-1}{-x - \sqrt{x^2-4x+1}}[/math][math] = \frac{-4x-1}{-x - \sqrt{x^2+4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

You can see this is false rather rapidly.

Okay this is based on Dapthar's original post. I orignally had a slightly different method in mind, but I also started out using the multiplication by the conjugate.

 

[math]\lim_{x\to\infty}{x+\sqrt{x^2-4x+1}[/math].

 

[math]x+\sqrt{x^2-4x+1} = (x+\sqrt{x^2-4x+1}) \cdot \frac{x - \sqrt{x^2-4x+1}}{x - \sqrt{x^2-4x+1}}(1)[/math]

 

[math] = \frac{x^2 - (x^2-4x+1)}{x - \sqrt{x^2-4x+1}} = \frac{4x-1}{x - \sqrt{x^2-4x+1}} = \frac{4x-1}{x - \sqrt{x^2-4x+1}}[/math][math]\cdot \frac{\frac{1}{x}}{\frac{1}{x}}[/math]

 

[math]=\frac{4+1/x}{1+\frac{1}{x} \sqrt{x^2+4x+1}}[/math]

 

let t = -1/x, such that [math] \lim_{x\to-\infty} = \lim_{t\to0} [/math]

 

[math]=\frac{4+t}{1+t \sqrt{\frac{1}{t^2}+\frac{4}{t}+1}}[/math]

[math]=\frac{4+t}{1+\sqrt{1+4t+t^2}}[/math]

 

By substituting t = 0, we get our desired 2.

From what I've seen, the methods are pretty much all the same. Dapthar's is probably the best because it doesn't involve any substitution.

That's not an error. Here i will show you what I'm referring to...

 

 

You can see this is false rather rapidly.

Yup, you're right. When I was going back through and changing all the signs I missed a few. I went back and edited my previous post, and fixed the signs again.

From what I've seen, the methods are pretty much all the same. Dapthar's is probably the best because it doesn't involve any substitution.

 

Actually, I just followed your proof, and it convinced me that the limit is 2. A good proof must be:

 

1. Utterly convincing

2. Short

 

I still have to look at Dapthar's. But I mean I remember that technique, multiplying by the conjugate, because the interior term vanishes.

 

Ok, I just followed Dapthar's as well, and both were convincing. I would say this... some folks will follow one better than the other, so I would do both.

Indeed. Always go for the way that you understand best. Comes out trumps every time (unless it's wrong ).

Indeed. Always go for the way that you understand best. Comes out trumps every time (unless it's wrong ).

 

Indeed.

[math] \lim_{x\to \infty} [-x + \sqrt{x^2+4x+1}] [/math]

 

Intuitively' date=' x^2 dominates over 4x. In other words, as x becomes increasingly huge, x^2 becomes enormous in comparison to 4x, so that we can neglect the addition of 4x to x^2, and certainly adding 1 doesn't alter the answer much either. Thus, we should expect the limit above to be equal to the following limit:

 

[math'] \lim_{x\to \infty} [-x + \sqrt{x^2}] = \lim_{x\to \infty} [-x + x] = \lim_{x\to \infty} [0] = 0 [/math]

 

Now, of course this is not rigorous, but it's intuitive.

 

And one other thing, the square root of x^2 is the absolute value of x.

I haven't seen anyone explain why this is wrong yet, even though everyone seems to agree that the limit is 2.

 

What you're doing is factoring out an x from the square root, so you get this:

 

[math] \lim_{x\to -\infty} [x + |x| \sqrt{1+\frac{4}{x}+\frac{1}{x^2}}] [/math]

 

So far so good. But then you're saying that what's left under the square root sign is going towards one as x goes to negative infinity, so that the limit above should go towards [math] \lim_{x\to -\infty} [x - x] = 0 [/math]. This would be true if x were finite -- but since x in this case is approaching infinity, multiplying it with something that's just approaching one doesn't necessarily leave it unchanged. It could converge to anything, or diverge, for all we know.