Infinity Minus Infinity Is Zero, Right?

Calculus Level 2

$\displaystyle\lim_{x\rightarrow \infty} \left( \sqrt{x^2+4x+1}-x\right) = \, ?$

The answer is 2.

3 solutions

Sravanth C.
Feb 23, 2016

Because the expression inside the limit has used a radical sign, we may consider solving this limit by conjugates.

The conjugate of the expression, $\sqrt{x^2 + 4x + 1} - x$ which is simply $\sqrt{x^2+4x+1} + x$ .

Let us multiply this expression, $\sqrt{x^2 + 4x + 1} - x$ by $\dfrac{\sqrt{x^2+4x+1} + x}{\sqrt{x^2+4x+1} + x}$ ,

$\begin{aligned} \displaystyle\lim_{x\rightarrow \infty} \left(\sqrt{x^2+4x+1}-x \right) &=&\displaystyle\lim_{x\rightarrow \infty} \left( \sqrt{x^2+4x+1}-x \right) \times \dfrac{\sqrt{x^2+4x+1}+x}{\sqrt{x^2+4x+1}+x}\\ &=&\displaystyle\lim_{x\rightarrow \infty} \dfrac{4x+1}{\sqrt{x^2+4x+1}+x}\\ &=&\displaystyle\lim_{x\rightarrow \infty} \dfrac{\frac{4x}{x}+\frac 1x}{\sqrt{\frac{x^2}{x^2}+\frac{4x}{x^2}+\frac 1{x^2}}+\frac{x}{x}}\\ &=&\dfrac{4+0}{\sqrt{1+0+0}+1} \\ &=&\boxed 2 \end{aligned}$

I used the same method

Leonardo Rodrigues - 5 years, 3 months ago

Guillermo Templado
Feb 23, 2016

$\lim_{x\to\infty} \sqrt{x^2 + 4x + 1} - x = \lim_{x\to\infty} \sqrt{(x + 2)^2 - 3} - x =$ $= \lim_{x\to\infty} \sqrt{(x + 2)^2 } - x = \lim_{x\to\infty} (x + 2) - x = 2$

Where did the -3 go

Abhinav Choudhary - 5 years, 3 months ago

When $x$ becomes unboundedly large, $(x+2)^2 - 3 \approx (x+2)^2$ .

Pi Han Goh - 5 years, 3 months ago

That is a very dangerous statement to say. It could be interpreted to imply that

$\lim_{x\rightarrow \infty} ( x+2)^3 - 3 - x^2 - 4x \approx \lim_{x\rightarrow \infty} ( x+2)^3 - x^2 - 4x ?$

What is the mathematically rigorous statement that you intended to communicate?

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – Oh right, it should be when $x\to \infty$ , the ratio $\dfrac{ (x+2)^2 - 3}{(x+2)^2} \to 1$ .

Pi Han Goh - 5 years, 3 months ago

@Pi Han Goh – Not quite.

$\lim \frac{ f(x) } { g(x) } = 1 \not \Rightarrow \lim f(x) - g(x) = 0.$

You seem to want to claim that

$\lim \frac{ f(x) } { g(x) } = 1 \not \Rightarrow \lim f(x) - h(x) = \lim g(x) - h(x).$

Do you think that is true?

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – Oh dang it. I bookmarked this page but I forgot to reply. Well, basically, I should add a square root sign to become: $\sqrt{(x+2)^2-3} \to x+2$ .

Thanks. It's always very tricky when dealing with infinities. I should be more careful next time~~

Pi Han Goh - 5 years, 3 months ago

@Calvin Lin – Don't you think in this case it is ok to do so? The largest degree will be x^3 anyway so chuck the 3. Can always make the russel paradox shit happen. Also i had a question. In Limit x tends to ∞ ((x-1)(x-2)(x-3)(x-4))^(1/4)-x can we use A.M. G.M. as all the multiplicating terms are linear and hence the infinities will be similar sowe use the A.M. as it is an upper bound.

akash omble - 5 years, 3 months ago

@Akash Omble –

No, I do not think it is okay to be mathematically ambiguous, unless you are explicit with your audience that you lack the necessary rigor. E.g. See the comments stream by Hummus below. There are additional conditions that need to hold, in order to justify the conclusion.
Note that there is no $x^3$ term.
I'm not sure how Russell paradox comes into play.
Yes, AM-GM can give an upper bound on that limit. However, I don't believe that it will provide the necessary lower bound (thought I might be wrong).

Calvin Lin Staff - 5 years, 3 months ago

same method!

Hamza A - 5 years, 3 months ago

I think the equation you deduced is wrong

Arunkumar Jayaraman - 5 years, 3 months ago

why?his solution contains no errors.although he approximated it,it's justified in this problem since the values approach infinity

Hamza A - 5 years, 3 months ago

The solution currently contains an unjustified statement. In particular, he does have to prove that $\lim_{x \rightarrow \infty} \sqrt{ (x+2)^2 - 3 } - (x+2) = 0$ .

Note that if $\lim f(x) = \infty$ and $\lim g(x) = \infty$ , it does not imply that $\lim f(x) - g(x) = 0$ . Hence, it does not follow immediately from "values approach infinity".

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – the derivative of the square root function gets infinitesimal as x approaches infinity

Hamza A - 5 years, 3 months ago

@Calvin Lin – what i meant was that $\lim _{ x\rightarrow \infty }{ \sqrt { (x+2)^{ 2 }-3 } } \approx x+2$

Hamza A - 5 years, 3 months ago

@Hamza A – The idea about derivatives is one way to prove my statement about the limits

Note that your limit equation doesn't make sense. The LHS is actually independent of $x$ , which is a dummy variable.

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – can you please elaborate on the limit part?how does it not make sense?

Hamza A - 5 years, 3 months ago

@Hamza A – What you meant to say is that

$\lim_{ x \rightarrow \infty } \sqrt{ (x+2) ^2 - 3 } - ( x + 2 ) = 0$

The reason why it doesn't your statement make sense, is because $x$ is a dummy variable when used in the limit, and could be replaced with anything. In particular,

$\lim_{ x \rightarrow \infty } \sqrt{ (x+2) ^2 - 3 } = \lim_{ N \rightarrow \infty} \sqrt{ (N+2)^2 - 3}$

Now, why would the RHS be (approximately) equal to $x + 2$ ?

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – because the derivative of the function $\sqrt{(x+2)^2-3}$ is never equal to zero(in the function's domain,x can be -2 but that will result in a negative number inside the square root) because the derivative is $\frac{x+2}{\sqrt{(x+2)^2-3}}$ ,it will get really close though,but the difference is still there,as x gets larger,the value of the derivative at that point(with the same x value) will get smaller and will approach 0 but never quite reach it

Hamza A - 5 years, 3 months ago

@Hamza A – I am not talking about the justification. As I said,

The idea about derivatives is one way to prove my statement about the limits.

Let me express the error that you have in another way.

Are the following statement true or false:
1. $\lim_{ N \rightarrow \infty} \sqrt{ (N+2)^2 - 3} = x + 2$ .
2. $\lim_{ x \rightarrow \infty} \sqrt{ (x+2)^2 - 3} = \lim_{ N \rightarrow \infty} \sqrt{ (N+2)^2 - 3}$ .

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – hmm,the first is true when x=N

the second is always right

Hamza A - 5 years, 3 months ago

@Hamza A – The first is not true. There is no "x" in the LHS, how can there be an "x" in the RHS?

The second statement is right, the first statement is wrong. And that is why it is not true that

$\lim _{ x\rightarrow \infty }{ \sqrt { (x+2)^{ 2 }-3 } } \approx x+2.$

That's all I'm saying about this.

Calvin Lin Staff - 5 years, 3 months ago

@Calvin Lin – oh,i get it,i over-complicated that for no reason :)

Hamza A - 5 years, 3 months ago

@Calvin Lin – I can prove that $\lim_{x \to \infty} \sqrt{(x+2)^2 - 3} - (x + 2) = 0 \Rightarrow$ $\lim_{x \to \infty} \sqrt{(x+2)^2 - 3} - x =\lim_{x \to \infty} \sqrt{(x+2)^2 - 3} - (x + 2) + \lim_{x \to \infty} (x+2) - x =$ $=\lim_{x \to \infty} (x+2) - x$ due to if there exits $\lim_{x\to a} f(x)$ and it's finite and there exits $\lim_{x\to a} g(x)$ and it's finite then there exists $\lim_{x\to a} f(x) + g(x) = \lim_{x\to a} f(x) + \lim_{x\to a} g(x)$ . And the end. Thank you anyway, I know now that this method is right...

Guillermo Templado - 5 years, 3 months ago

Muhammad Maulana
Feb 29, 2016

Another $\text{Method 1:}$ $\boxed{\displaystyle\large\lim_{x\rightarrow\infty} \sqrt{(a x^{2}+bx+c)} - \sqrt{(a x^{2}+px+q)}= \frac {b-p}{2 \sqrt a}}$ $\displaystyle \large \lim_{x\rightarrow\infty} \sqrt{(x^{2}+4x+1)} -x = \displaystyle\large\lim_{x\rightarrow\infty} \sqrt{(x^{2}+4x+1)} - \sqrt{x^{2}}$ $\displaystyle\large\lim_{x\rightarrow\infty} \sqrt{(x^{2}+4x+1)} - \sqrt{x^{2}}$ Subsitued $\boxed{a=1,b=4,p=0}$ : $\displaystyle\large\lim_{x\rightarrow\infty} \sqrt{(x^{2}+4x+1)} - \sqrt{x^{2}} = \frac {b-p}{2 \sqrt{a}} =\frac {4-0}{2 \sqrt1}=\boxed {2}$ $\text{Method 2:}$ $\boxed{\displaystyle\large\lim_{x\rightarrow\infty} (ax+b) - \sqrt{(a^{2} x^{2}+px+q)}= \frac {2ab-p}{2a}}$ $\displaystyle \large \lim_{x\rightarrow\infty} \sqrt{(x^{2}+4x+1)} -x = \displaystyle\large -\lim_{x\rightarrow\infty} x- \sqrt{(x^{2}+4x+1)}$ Subsitued $\boxed{a=1,b=0,p=4}$ : $\displaystyle\large -\lim_{x\rightarrow\infty} x-\sqrt{(x^{2}+4x+1)} = - \frac {(2ab-p)}{2a} = -\frac {(2.1.0-4)}2=\boxed {-\frac{(-4)}2 = 2}$

Infinity Minus Infinity Is Zero, Right?

The answer is 2.

3 solutions

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