Why can't I solve the limit by factoring out the variable here? 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? 
 A: As per the comment, because $0\cdot \infty$ is indeterminate form.

Your second limit is 
$$\lim_{n\rightarrow\infty}\left(n - \sqrt{n + a}\sqrt{n + b}\right)=
\lim_{n\rightarrow\infty}\frac{n^2 - (n + a)(n + b)}{n + \sqrt{n + a}\sqrt{n + b}}=\\
\lim_{n\rightarrow\infty}\frac{-(a+b)n-ab}{n + \sqrt{n + a}\sqrt{n + b}}=\\
\lim_{n\rightarrow\infty}\frac{-(a+b)-\frac{ab}{n}}{1 + \sqrt{1 + \frac{a}{n}}\sqrt{1 + \frac{b}{n}}}=-\frac{a+b}{2}$$

Whether the 1st one, just to have an alternative proof using binomial coefficients (also covered here and partially here)
$$|c|=\frac{1}{1+b}<1 \Rightarrow b>0$$
and $(1+b)^n \geq 1+nb+\frac{n(n-1)}{2}b^2, n\geq 2$. Thus
$$0<\lim\limits_{n\to\infty}n|c|^n < \lim\limits_{n\to\infty}\frac{n}{1+nb+\frac{n(n-1)}{2}b^2}\to 0, n\to\infty$$
A: The other responders are correct that $0 \cdot \infty$ is an indeterminate form.  But it seems to me it is also important to point out that the original poster's answer for the limit $\lim_{x \to \infty} x c^x$ is written in a confusing way.  In particular, the step
$$
\lim_{x \to \infty} x \left(\frac{\log x}{x} + \log c \right) = \lim_{x \to \infty} x \log c
$$
is misleading, because it seems to suggest that you can compute the limit of a product by filling in the limit of first one factor and then the other.  A better way to compute this limit would be to compute the limits of the two factors separately: $$\lim_{x \to \infty} x = \infty, \qquad \lim_{x \to \infty} \left(\frac{\log x}{x} + \log c\right) = \log c.$$  Now you can appeal to a general theorem that says that if $\lim f(x) = \infty$ and $\lim g(x) = M < 0$, then $\lim f(x) \cdot g(x) = -\infty$.  To say that $0 \cdot \infty$ is an indeterminate form simply means that there is no similar theorem for the case $\lim f(x) = \infty$ and $\lim g(x) = 0$.
While it is helpful to point out what theorems don't exist (which is what other responders are doing when they say that $0 \cdot \infty$ is indeterminate), one really needs to know what theorems do exist.  For a general theorem about limits of products where one or both factors approach $\pm\infty$, see:  Velleman, Calculus: A Rigorous First Course, Theorem 2.5.14, pp. 92-93.
By the way, a small point:  The solution for $\lim_{x \to \infty} xc^x$ is correct for $0 < c < 1$.  The original question was about $\lim_{n \to \infty} n c^n$ for $|c| < 1$, so one should give a separate argument for the cases $-1 < c < 0$ and $c=0$.
A: The problem with your product is that you have $$\infty \times 0$$ 
Which is indeterminate so the answer is not necessarily $0$ 
