When is a limit of products not a product of limits? Suppose I have the following limiting expression:
$$\alpha = \lim_{x\to \infty} (f(x) - x^2 g(x))<\infty,$$
with $f,g>0$ for all $x$. Furthermore, suppose $f\to\infty$ and $x^2g\to\infty$ as $x\to\infty$, but that $\lim_{x \rightarrow \infty} xg(x)=c<\infty$. Can I always replace $xg(x)$ with $c$, e.g. will it always be legal to write
$$\alpha=\lim_{x\to\infty}(f(x)-xc)\quad?$$
 A: If $g(x)\rightarrow \infty$ as $x\rightarrow \infty$, then $xg(x)\rightarrow \infty$ as $x\rightarrow \infty$; see here for a proof. In other words $\lim\limits_{x\rightarrow \infty}{xg(x)}$ simply cannot exist.
A: I think you want to assume $x^2g(x) \rightarrow \infty$ as $x \rightarrow \infty$ here and  not $g(x) \rightarrow \infty$. 
Still the statement doesn't have to hold. Let $g(x) =  cx^{-1} + x^{-{3 \over 2}}$ and 
$f(x) = x^2g(x) + \alpha$. 
Then automatically $\lim_{x\to \infty} (f(x) - x^2 g(x)) = \alpha$ and also $\lim_{x \rightarrow \infty} xg(x) = c$
Since $f(x) - xc = x^2g(x) + \alpha - xc = x^2(cx^{-1} + x^{-{3 \over 2}}) + \alpha - xc = x^{1 \over 2} + \alpha$, you have $\lim_{x \rightarrow \infty} f(x) - xc = \infty$ and not $\alpha$.
A: While evaluating limits of a complicated expression consisting of many subexpressions connected by $+, -, \times, /$, one should not replace a subexpression by its limit. This is plain wrong and the rules of algebra of limits do not allow this. However it seems that it is a common approach followed by students with great success.
The reason this approach works is that there are two specific instances in which this approach is justified.
Let $E$ be a complicated expression whose limit we wish to calculate. And let $B$ be one subexpression of $E$. Let everything else in $E$ apart from $B$ be called $A$. Then we can replace subexpression $B$ by its limit $b$ if either
1) $E = A \pm B$
or
2) $E = A \cdot B$ (in this case $b$ must be non-zero)
Thus the subexpression which we want to replace with its limit must be either connected via $\pm$ or $\times$ with the rest of expression as the other operand of $\pm$ or $\times$. Also the limit of the subexpression must be  non-zero if it is connected by $\times$.
In the current question the expression $E = f(x) - x^{2}g(x)$ and $B = xg(x)$ and limit of $B$ is $c$. We can clearly observe that here we don't have $E = A \pm B$ or $E = A \cdot B$ for some subexpression $A$. Hence we can't replace $B = xg(x)$ by its limit $c$.
This topic has been explained with examples in my blog post under the heading "Misuse of Rules of Limits". 
A: No, it's not legal. For example, take $f(x)=x$ and $g(x)=\frac{1}{x}-\frac{1}{x^2}$. Then $x^2g(x)=x-1$, so
$$
\lim_{x\to\infty} (f(x)-x^2g(x))=\lim_{x\to\infty}(x-x+1)=1.
$$
And $xg(x)=1-\frac{1}{x}$, so $c=1$. But
$$
\lim_{x\to\infty}(f(x)-xc)=0.
$$
When you replace $x^2g(x)$ with $xc$ inside the limits, the error is in
$$
\lim_{x\to\infty}x^2 g(x)= x\lim_{x\to\infty} xg(x)= xc.
$$
