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.

  • $\begingroup$ 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

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 $x\to\infty$.)

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

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