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?

  • $\begingroup$ Because it's a $0 \cdot \infty$ case, aka indeterminate form. $\endgroup$ – rtybase Aug 16 '19 at 21:38
  • $\begingroup$ The difference is that direct substitution giving $\infty\times k$ where $k\ne0$ is NOT an indeterminate form. If we instead have $\infty\times0$ this is an indeterminate form and so can take any value (consider $n\times \frac{a}{n}$). $\endgroup$ – Peter Foreman Aug 16 '19 at 21:38
  • $\begingroup$ $\lim\limits _{n \rightarrow \infty}n - \sqrt{n+a}\sqrt{n+b} = \lim\limits_{n \rightarrow \infty} n - \sqrt{n^2 + n(a + b) + ab} = \lim\limits_{n \rightarrow \infty} n\left(1 - \sqrt{1 + \frac{1}{n}(a+b) + \frac{ab}{n^2}}\right)=\lim_{n \rightarrow \infty}n\left(1 - \left(1+\frac1{2n}(a+b)\dots\right)\right)=\dfrac{-1}2(a+b)$ $\endgroup$ – J. W. Tanner Aug 16 '19 at 21:38
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    $\begingroup$ @rtybase oof. here i thought $0 \cdot \infty$ wasn't indeterminate. Thanks! (if you make your comment an answer, I'll accept it) $\endgroup$ – AmagicalFishy Aug 16 '19 at 21:41
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    $\begingroup$ binomial series: $(1+x)^{1/2}=1+\dfrac x2\dots$ $\endgroup$ – J. W. Tanner Aug 16 '19 at 21:46

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 $$0<\lim\limits_{n\to\infty}n|c|^n < \lim\limits_{n\to\infty}\frac{n}{1+nb+\frac{n(n-1)}{2}b^2}\to 0, n\to\infty$$


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 $c=0$.


The problem with your product is that you have $$\infty \times 0$$

Which is indeterminate so the answer is not necessarily $0$


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