0
$\begingroup$

Consider the limit...

$$\lim_{x\rightarrow\infty} \left( x^2 \sin{\frac{1}{x^2 - 1} - \sin{\frac{1}{x^2-1}}}\right)$$

First, the correct way of solving it: rewrite the limit as... $$\lim_{x\rightarrow\infty}\frac{\sin{\frac{1}{x^2-1}}}{\frac{1}{x^2-1}}$$ ... and note that, if $u = \frac{1}{x^2 - 1} \rightarrow 0$ as $x \rightarrow \infty$. Thus, we re-write the above limit as... $$\lim_{u\rightarrow0} \frac{\sin{u}}{u} = 1$$

Ok. Now the wrong way, doing what is done for many other problems where $x\rightarrow\infty$. Since $\frac{1}{x^2 - 1} \rightarrow 0$ when $x \rightarrow \infty$, we have...

$$\infty \cdot \sin{0} - \sin{0} = \infty$$

This is the incorrect answer—but why does the strategy of plugging in $\infty$ suddenly fail? What mathematics does this strategy depend on such that it doesn't work for this problem, but works for this one?

$$\lim_{x\rightarrow\infty} \left( x^3 \sin{\frac{1}{x^2 - 1} - \sin{\frac{1}{x^2-1}}}\right)$$

It's clear that the $x^3$ grows at a much faster rate, thus breaking out of whatever ceiling is created by the $\sin$, but, outside of qualitative descriptions of what's going on, why does plopping $\infty$ as if it were a number work for this, and so many other relatively simple limits, but not the one presented above (with $x^2$)?

Edit: I'm aware that $\infty$ isn't a number and saying something like $\frac{1}{\infty}$ is a convenient, but technically incorrect notation.

$\endgroup$
  • $\begingroup$ I am not suer we can write $\infty \cdot \sin{0} - \sin{0} = \infty$. From a philosophy point of view, if we use a concept in an invalid way, and it yields a correct answer, we could comfortably assume that it is a coincidence. In limits, you can get correct answers by not applying the rules sometimes. $\endgroup$ – NoChance Apr 27 at 1:13
  • $\begingroup$ "Plugging in $\infty$" has never worked, in the sense that it's mathematical nonsense. You can learn some mnemonics for some valid mathematical theorems involving limits to infinity, e.g. $1/\infty = 0$ is a mnemonic for the valid mathematical theorem that if $f(x) \to 1$ and $g(x) \to \infty$, then $f(x) / g(x) \to 0$, but the statement $1 / \infty = 0$ is mathematically meaningless. $\endgroup$ – Theo Bendit Apr 27 at 1:13
  • $\begingroup$ There are plenty of examples of the $0 \cdot \infty$ mnemonic breaking down. For example, consider $\frac{1}{x} \cdot x, \frac{1}{x^2} \cdot x, \frac{1}{x} \cdot x^2$. $\endgroup$ – Theo Bendit Apr 27 at 1:15
  • $\begingroup$ $\infty\cdot \sin 0 = \infty \cdot 0$, which is an indeterminate form, not $\infty$. $\endgroup$ – Greg Martin Apr 27 at 1:25
  • $\begingroup$ @GregMartin That's what I was missing—huuuge oversite. If you make your comment an answer, I'll accept it. $\endgroup$ – AmagicalFishy Apr 27 at 1:26
2
$\begingroup$

The error in reasoning is that $\infty\cdot \sin 0 = \infty \cdot 0$, which is an indeterminate form, not a result of $\infty$. (And as Theo Bendit commented, the reason it is an indeterminate form is because of examples like the limits of $$ x \cdot \frac1x,\quad 2x \cdot \frac1x,\quad 0.3x \cdot \frac1x,\quad x^2 \cdot \frac1x,\quad x \cdot \frac1{x^2} $$ as $x\to\infty$.)

$\endgroup$
  • $\begingroup$ For some reason, I was telling myself $\sin{0} = 1$ $\endgroup$ – AmagicalFishy Apr 27 at 1:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.