This is from Spivak's Calculus, Chp. 22, Q2 (vi): Consider $\lim_{n\rightarrow\infty} nc^n, |c| < 1$...
$$\lim_{x \rightarrow \infty} xc^x = \lim_{x \rightarrow \infty} e^{\log{x}} e^{x\log{c}} = \lim_{x \rightarrow \infty} e^{\log{x} + x\log{c}}$$
... then...
$$\lim_{x \rightarrow \infty} \log{x} + n\log{c} = \lim_{x \rightarrow \infty} x\left( \frac{\log{x}}{x} + \log{c}\right) = \lim_{x \rightarrow \infty} x \log{c} = -\infty$$
So $\lim_{x \rightarrow \infty} xc^x = 0$. In particular, $\lim_{n \rightarrow \infty}nc^n = 0$. Notice that he used (I think) the fact that the limit of a product is the product of the limits in the last step (to solve one limit independently and reduce the limit to two logs). I thought this only worked if each limit in the product was finite, but I guess it's usable here?
Now, consider this limit from Q2 (ii):
$\lim_{n\rightarrow\infty}n - \sqrt{n + a}\sqrt{n + b}$. The correct way to evaluate this limit requires multiplying by $\frac{n + \sqrt{n+a}\sqrt{n+b}}{n + \sqrt{n+a}\sqrt{n+b}}$, which ultimately leads to a limit of $-\frac{a+b}{2}$.
But, why can't I do this (which leads to the incorrect answer):
$$\lim_{n \rightarrow \infty}n - \sqrt{n+a}\sqrt{n+b} = \lim_{n \rightarrow \infty} n - \sqrt{n^2 + n(a + b) + ab} = \lim_{n \rightarrow \infty} n\left(1 - \sqrt{1 + \frac{1}{n}(a+b) + \frac{ab}{n^2}}\right)=\lim_{n \rightarrow \infty}n(1 - \sqrt{1}) = 0$$
This seems to coincide well with the taking-the-limit-of-one-term used in the first presented problem (one of the limits is finite, the other is not). Why does this one fail?