Sweet and Sour Limits

Eagle-eyed friend of the blog @robjlow spotted an error in Uncle Colin’s last answer. As I’m forever telling my students, making errors is how you learn; Rob has graciously delivered a lesson for us all. Thanks for keeping me honest!

Recently, Uncle Colin gave a couple of ways to see that
\[ \lim_{x \to \infty} \left\{ \sqrt{x^2 + 3x} – x\right\} = \frac{3}{2} \]

One approach, the sweet, used the binomial theorem, to write
\[ \sqrt{x^2 + 3x} = x\sqrt{1+\frac{3}{x}} \approx x\left(1+\frac{3}{2x}\right). \]

The very suspicious minded mind be a little hesitant, because the error in the binomial approximation is multiplied by \(x\), which is large. But we know that subsequent terms in the binomial expansion are multiples of higher negative powers of \(x\), so that’s fairly obviously OK. (Which isn’t to say that we don’t have to check!)

The second approach is the sour. Here, Colin factorizes to get
\[ \sqrt{x^2 + 3x} = \left(x+\frac{3}{2}\right) \sqrt{1-\frac{9}{4\left(x+\frac{3}{2}\right)^2}} \]
Then the argument is that as \(x \to \infty\), the square root term tends to 1, and we’re just left with \(x+\frac{3}{2}-x = \frac{3}{2}\).

So what’s the problem?

It’s the same problem as my suspicious mind had in the first approach, but now without the obvious fix. Now all we have is that the square root term tends to 1, and I’ve no idea of what the error is.

Of course, the final result is correct, so if I want to convince you that there is a problem I need an example which follows exactly the same reasoning, but gives a wrong answer.

Consider
\[ \lim_{x \to \infty} \left\{ x \sqrt{1+\frac{1}{x}} – x \right\}. \]
This has a very similar form to the original problem.
The square root term still tends to 1 as \(x \to \infty\), leaving \(x-x = 0\) – but this time we don’t get the right limit.

Of course, to persuade you of that, I need to do more than just make an assertion, I have to calculate the limit convincingly (and get a different answer).

I’m going to do it using a method which is less general than the binomial expansion, but leaves no lingering aftertaste of “was my approximation good enough?”. This is a useful trick when you have a difference of terms involving square roots of large quantities, and consists of multiplying the expression by 1, but expressed in a useful way.

\[ \begin{split}
x \sqrt{1+\frac{1}{x}} – x &=
\left( x \sqrt{1+\frac{1}{x}} – x \right) \frac{x \sqrt{1+\frac{1}{x}} + x}{x \sqrt{1+\frac{1}{x}} + x}\\
&= \frac{x^2\left(1+\frac{1}{x}\right) – x^2}{x \sqrt{1+\frac{1}{x}} + x}
\\ &= \frac{x}{x \sqrt{1+\frac{1}{x}} + x} \\
&= \frac{1}{\sqrt{1+\frac{1}{x}} + 1}
\end{split}
\]
From this, it’s obvious that
\[
\lim_{x \to \infty} \left\{ x \sqrt{1+\frac{1}{x}} – x \right\}=\frac{1}{2}.
\]
If you work it out using the binomial approximation, you’ll see that you get the same answer that way.

So what soured the second approach?

In this case, the error involved in using \(\sqrt{1-1/x} \approx 1\) was small, but the error has to be multiplied by \(x\), which is large. We need the error to be so small that it is still small after multiplication by \(x\), and this time it isn’t.

It’s all a bit subtle: Colin’s answer was right, it was the argument he used to get it that was wrong, and there’s nothing to stop a wrong argument from giving the right answer. If you think you have a proof of something, you have to make sure that the steps in the argument work as advertised. This might involve filling in detail (like getting an approximation for the error and making sure it is small enough), or it might involve just using a different type of argument.

It might seem a bit sad that checking you have the right answer doesn’t also check that you have a correct solution, but that’s the kind of thing that makes mathematics fun.

Robert Low

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