Computing a limit involving a square root: what is wrong?

[yucheng]

Homework Statement

> Rudin 3.4. Calculate $\lim\limits_{n \to \infty} \left(\sqrt{n^2 + n} - n\right)$.

Relevant Equations

N/A

My attempt:

\begin{align}
\lim\limits_{n \to \infty} \sqrt{n^2 + n} - n &= n\sqrt{1+\frac{1}{n}} -n\\
&=n - n\\
&= 0\\
\end{align}

I think the issue is at (1)-(2)

For comparison, here is Rudin's solution

 

[Orodruin]

You are missing one term that does not vanish in (2). I suggest you expand the square root in a Taylor expansion in $1/n$.

Edit: The issue is that, by just taking the leading term of the square root, you end up with $n\sqrt{1 + 1/n} = n(1 + \mathcal O(1/n)) = n + \mathcal O(1)$. The $\mathcal O(1)$ term does not vanish as $n\to \infty$ and so just taking the leading term from the square root tells you nothing about the limit.

 

[FactChecker]

Although $\sqrt{1+1/n}$ gets arbitrarily close to 1, it is then multiplied by $n$, which is large. So it is hard to conclude that $n\sqrt{1+1/n}$ approaches $n$.

 

[PeroK]

Homework Statement:: > Rudin 3.4. Calculate $\lim\limits_{n \to \infty} \left(\sqrt{n^2 + n} - n\right)$.
Relevant Equations:: N/A

My attempt:

\begin{align}
\lim\limits_{n \to \infty} \sqrt{n^2 + n} - n &= n\sqrt{1+\frac{1}{n}} -n\\
&=n - n\\
&= 0\\
\end{align}

I think the issue is at (1)-(2)

For comparison, here is Rudin's solution

Another approach is to complete the square for $n^2 + n$.

 

[yucheng]

@PeroK

\begin{align*}
\lim\limits_{n \to \infty} \sqrt{n^2 + n} - n &= \lim\limits_{n \to \infty} \sqrt{(n+\frac{1}{2})^2 -\frac{1}{4}} - n \\
&= \lim\limits_{n \to \infty} n+\frac{1}{2} - n \\
&= \frac{1}{2} \\
\end{align*}

 

[PeroK]

@PeroK

\begin{align*}
\lim\limits_{n \to \infty} \sqrt{n^2 + n} - n &= \lim\limits_{n \to \infty} \sqrt{(n+\frac{1}{2})^2 -\frac{1}{4}} - n \\
&= \lim\limits_{n \to \infty} n+\frac{1}{2} - n \\
&= \frac{1}{2} \\
\end{align*}

That's not a rigorous proof, but it does clarify the answer. You can, of course, do something rigorous with the first term.

 

[yucheng]

That's not a rigorous proof, but it does clarify the answer. You can, of course, do something rigorous with the first term.

I don't think I know how to rigorize this. Do you mean:

\begin{align*}
\lim\limits_{n \to \infty} \sqrt{(n+\frac{1}{2})^2 -\frac{1}{4}} &= \lim\limits_{n \to \infty} (n+\frac{1}{2})\sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}} \\
&= \lim\limits_{n \to \infty} n+\frac{1}{2}
\end{align*}

Since $(n+\frac{1}{2})\sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}} = (n+\frac{1}{2})(1 + \mathcal O(\frac{1}{(n+\frac{1}{2})^2)}) = (n+\frac{1}{2}) + \mathcal O(\frac{1}{(n+\frac{1}{2}))})$

P.S. i just plugged and chugged and @Orodruin 's argument :)

 

[Orodruin]

I don't think I know how to rigorize this. Do you mean:

\begin{align*}
\lim\limits_{n \to \infty} \sqrt{(n+\frac{1}{2})^2 -\frac{1}{4}} &= \lim\limits_{n \to \infty} (n+\frac{1}{2})\sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}} \\
&= \lim\limits_{n \to \infty} n+\frac{1}{2}
\end{align*}

Since $(n+\frac{1}{2})\sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}} = (n+\frac{1}{2})(1 + \mathcal O(\frac{1}{(n+\frac{1}{2})^2)}) = (n+\frac{1}{2}) + \mathcal O(\frac{1}{(n+\frac{1}{2}))})$

P.S. i just plugged and chugged and @Orodruin 's argument :)

That's not how I would do it though. I would have started from your $n\sqrt{1 + 1/n}$ and noted that $\sqrt{1 + x} = 1 + x/2 + \mathcal O(x^2)$ to find
$$
n\sqrt{1 + \frac 1n}-n = n \left[ 1 + \frac{1}{2n} + \mathcal O(1/n^2)\right]-n = n + \frac 12 + \mathcal O(1/n) -n = \frac 12 + \mathcal O(1/n).
$$

 

[Office_Shredder]

You keep writing all these limits and claiming they equal each other. They do, in as far as they all are equal to infinity in your last post, but that actually doesn't help you solve your problem. I think you should try to be more careful with what you write, since what you're doing is not going to do you any favors as you progress in your studies.

 

[Mark44]

I don't think I know how to rigorize this. Do you mean:

\begin{align*}
\lim\limits_{n \to \infty} \sqrt{(n+\frac{1}{2})^2 -\frac{1}{4}} &= \lim\limits_{n \to \infty} (n+\frac{1}{2})\sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}} \\
&= \lim\limits_{n \to \infty} n+\frac{1}{2}
\end{align*}

This is incorrect, as the last limit you get is $\infty$. The value I get for this limit is $\frac 1 2$.
Probably the easiest approach is to multiply the first expression by 1, in the form of $\sqrt{n^2 + n} + n$ over itself. When you have a limit that involves the sum or difference of square root quantities, it's often helpful to multiply by the conjugate over itsef.

Also, I don't think that the approach shown by @Orodruin, (with the "big O" notation) is necessary here, and is not an approach that is normally taken in most calculus textbooks.

 

[Mark44]

Another tip to make things easier in limit problems is use algebraic manipulation to get the expression into some usable form, and then take the limit.

For example, do the algebraic manipulation:
$$\sqrt{n^2 + n} - n = (\sqrt{n^2 + n} - n) \cdot \frac{\sqrt{n^2 + n} + n}{\sqrt{n^2 + n} + n} = \dots = \text{final expression}$$
Now take the limit:
$$ \lim_{n \to \infty} \sqrt{n^2 + n} - n = \lim_{n \to \infty} \text{final expression} = L$$

 

[Orodruin]

Also, I don't think that the approach shown by @Orodruin, (with the "big O" notation) is necessary here, and is not an approach that is normally taken in most calculus textbooks.

It’s just how I would do it to get the limit. As long as you get the limit correct, nothing else is really necessary.

 

[vela]

\begin{align*}
\lim_{n \to \infty} \sqrt{(n+\frac{1}{2})^2 -\frac{1}{4}} &= \lim_{n \to \infty} (n+\frac{1}{2})\sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}} \\
&= \lim_{n \to \infty} n+\frac{1}{2}
\end{align*}

You seem to be using this line of reasoning:
\begin{align*}
\lim_{n \to \infty} \sqrt{(n+\frac{1}{2})^2 -\frac{1}{4}}
&= \lim_{n \to \infty} (n+\frac{1}{2})\sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}} \\
&= \lim_{n \to \infty} (n+\frac{1}{2}) \underbrace{\lim_{n \to \infty} \sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}}}_1 \\
&= \lim_{n \to \infty} n+\frac{1}{2}
\end{align*} That second line is not correct, and the same mistake led to the erroneous result in your original post.

 

[pasmith]

Alternatively, $$\begin{align*}<br /> \sqrt{n^2 + n} - n &= n\left(\sqrt{1 + \frac1n} - 1\right) \\<br /> &= \frac{\sqrt{1 + \frac1n} - 1}{\frac 1n} <br /> \end{align*}$$ and hence by l'Hopital's rule $$\begin{align*}<br /> \lim_{n \to \infty} \sqrt{n^2 + n} - n &= \lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x} \\<br /> &= \lim_{x \to 0} \frac{\frac{1}{2\sqrt{1 + x}}}{1} = \frac12.\end{align*}$$

 

[PeroK]

I don't think I know how to rigorize this. Do you mean:

\begin{align*}
\lim\limits_{n \to \infty} \sqrt{(n+\frac{1}{2})^2 -\frac{1}{4}} &= \lim\limits_{n \to \infty} (n+\frac{1}{2})\sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}} \\
&= \lim\limits_{n \to \infty} n+\frac{1}{2}
\end{align*}

Since $(n+\frac{1}{2})\sqrt{ 1 - \frac{1}{4(n+\frac{1}{2})^2}} = (n+\frac{1}{2})(1 + \mathcal O(\frac{1}{(n+\frac{1}{2})^2)}) = (n+\frac{1}{2}) + \mathcal O(\frac{1}{(n+\frac{1}{2}))})$

P.S. i just plugged and chugged and @Orodruin 's argument :)

You need to show that as $n$ increases, so $\sqrt{(n+\frac 1 2)^2 - \frac 1 4}$ gets arbitrarily close to $n + \frac 1 2$. We can see that it is always less than $n + \frac 1 2$, so we need to show that for every $\epsilon > 0$ we can find $N$ such that $n > N \ \Rightarrow \sqrt{(n+\frac 1 2)^2 - \frac 1 4} > (n + \frac 1 2) - \epsilon$.

One technique is to work forwards to find $N$, then work backwards from that in your proof. In this case, we need $n + \frac 1 2 > \frac 1 2(\epsilon + \frac 1 {4\epsilon})$. It would be a good exercise for you to show that that works and then to work out for yourself how to find that lower bound.

Other methods are simpler for this particular problem, but if you are doing real analysis you need to be able to do these $\epsilon$ based proofs.