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.
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 $x\to\infty$.)