Limits and infinity minus infinity

Question

I understand that we need to rationalize when we have infinity minus infinity like here

$\lim_{x\to \infty}\left(\sqrt{x^2 + 1} - \sqrt{x^2 + 2}\right)$

My question is why can I not just split the limits like this

$\lim_{x\to \infty}\left(\sqrt{x^2 + 1}\right) - \lim_{x\to \infty}\left(\sqrt{x^2 + 2}\right)$

and then

$\lim_{x\to \infty}\sqrt{x^2} * \lim_{x\to \infty}\sqrt{1 + \frac 1x} - \lim_{x\to \infty}\sqrt{x^2} * \lim_{x\to \infty}\sqrt{1 + \frac 2x}$

which gives

$\lim_{x\to \infty}\sqrt{x^2} - \lim_{x\to \infty}\sqrt{x^2} = 0$

because $\frac 1x$ and $\frac 2x$ tend to $0$

Where am I wrong?

Answer 1

Infinity is not a number.

So there there is no general meaning to any "infinity arithmetic" expression.

Sometimes, though, there is a limit theorem which can be interpreted as an infinity arithmetic expression.

Here's one example of such a theorem:

Theorem: Given sequences $(x_n)$ and $(y_n)$ in $\mathbb R$, if $\lim_{n \to \infty} x_n = \infty$, and if $\lim_{n \to \infty} y_n = \infty$, then $\lim_{n \to \infty} (x_n + y_n) = \infty$.

Because of this theorem, one might argue that it is fair to "split the limits", as you say, resulting in the "infinity arithmetic" expression. $$\infty + \infty = \infty $$ Fine so far.

But just because one can write an "infinity arithmetic" expression does not mean there is a theorem supporting that expression.

So, for example, there is NO theorem like this:

False Theorem: Given sequences $(x_n)$ and $(y_n)$ in $\mathbb R$, if $\lim_{n\to\infty} x_n = \infty$ and $\lim_{n \to \infty} y_n = \infty$ then $\lim_{n \to \infty} (x_n - y_n) = $ BLAH.

It doesn't matter what you substitute for BLAH, the resulting statement will be false. Whether you substitute BLAH $=0$, or BLAH $= 1$, or BLAH $=42$ or BLAH $=$anything else, the resulting statement will be false.

To prove this, let me give you two counterexamples:

• Counterexample 1: If $x_n = n$ and $y_n = n$ then $\lim_{n\to\infty} x_n = \infty$, and $\lim_{n \to \infty} y_n = \infty$, and $$\lim_{n \to \infty} (x_n - y_n)= \lim_{n \to \infty} (n - n) = \lim_{n \to \infty} 0 = 0 $$
• Counterexample 2: If $x_n = n$ and $y_n = n-1$ then $\lim_{n \to \infty} x_n = \infty$ and $\lim_{n \to \infty} y_n$ and $$\lim_{n \to \infty}(x_n-y_n) = \lim_{n \to \infty} (n - (n-1)) = \lim_{n \to \infty} 1 = 1 $$

So, if you tried to convince me that the "False Theorem" was true using the substitution BLAH $=0$, I would show you Counterexample 2. And if you tried to convince me that the "False Theorem" was true using any substitution not equal to $0$, such as BLAH $=1$ or BLAH $=42$ or BLAH $=\infty$ or BLAH $=$anything else not equal to zero, then I would show you Counterexample 1.

In Calculus 1 we teach that the expression "$\infty-\infty$" is an indeterminate form. What this really means is what I've said above: there is no limit theorem which justifies any evaluation of $\infty-\infty$. When you encounter what looks like an $\infty-\infty$ expression, your best mathematical strategy is to DO SOMETHING ELSE, i.e. to re-evaluate the expression, rewrite it, alter it in some fashion (obeying the laws of algebra), so that it no longer has the $\infty-\infty$ form.

So for the opening example in your post, the mathematical strategy, as you say, is to rationalize the radicals. And you have probably learned other $\infty-\infty$ examples with different mathematical strategies.

Answer 2

You can not subtract $$ \lim_{x\to\infty}(\sqrt{x^2+1}-\sqrt{x^2+2})=\lim_{x\to\infty}\sqrt{x^2+1}-\lim_{x\to\infty}\sqrt{x^2+2}, $$ because the left hand side is $\infty-\infty$ which is not definable.

Instead, you can obtain that $$ \sqrt{x^2+1}-\sqrt{x^2+2}=\frac{(\sqrt{x^2+1}-\sqrt{x^2+2})(\sqrt{x^2+1}+\sqrt{x^2+2})}{\sqrt{x^2+1}+\sqrt{x^2+2}}\\ =\frac{(x^2+1)-(x^2+2)}{\sqrt{x^2+1}+\sqrt{x^2+2}}=-\frac{1}{\sqrt{x^2+1}+\sqrt{x^2+2}}\to 0, $$ where the properties of limits apply.

Answer 3

As others have pointed out, the fundamental flaw in the OP is in the assumption that $\infty-\infty$ has a meaning. This perhaps becomes more clear if we note that $\sqrt{x^2}\sqrt{1+{1\over x}}=\sqrt{x^2+x}$, not $\sqrt{x^2+1}$, so the incorrect computation in the OP is actually for $\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2+2x})$, for which a correct approach gives

$$\lim_{x\to\infty}(\sqrt{x^2+x}-\sqrt{x^2+2x})=\lim_{x\to\infty}{-x\over\sqrt{x^2+x}+\sqrt{x^2+2x}}=\lim_{x\to\infty}{-1\over\sqrt{1+{1\over x}}+\sqrt{1+{2\over x}}}=-{1\over2}$$

and not $0$.