calculus - $lim_{ntoinfty}a_nb_n=? $ given that $lim_{ntoinfty}a_n=0$ and $lim_{ntoinfty}b_n=infty$

A. $\lim_{n\to\infty}a_nb_n=?$ given that $\lim_{n\to\infty}a_n=0$ and $ \lim_{n\to\infty}b_n=\infty$

B. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}b_n=0$

C. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=\infty$ and $\lim_{n\to\infty}b_n= \infty$

These seem to be difficult situations I come across when I have to evaluate limits.

Some examples,

1. $\lim_{n\to\infty}n^a\sin(1/n^b)$ with $a,b>1$


2. $\lim_{n\to\infty}{n^a\over \sin(1/n^b)}$ with $a,b>1$


3. $\lim_{n\to\infty}{1/n^a\over \sin(1/n^b)}$ with $a,b>1$


4. $\lim_{n\to\infty} nf(x_n)$ [where we only know that $\lim \ f(x_n) = 0]$

My first question is : Are the two situations A and B equivalent ? I feel that they are equivalent as $\lim_{n\to\infty}b_n=\infty \iff \lim_{n\to\infty}{1\over b_n}=0$, The problem is that I'm not sure if we can apply the product rule for limits to take the limit inside.

My second question is : What are some ways to solve such problems ? I only know that we can apply stolz- cesaro in some cases.

Answer

These are usually called indeterminate forms, not because we can't compute the limit, but because no general theorem exists.

Just to make a couple of easy examples for your fourth question.

Consider $b_n=\frac{1}{n}$. Then
$$
\lim_{n\to\infty} b_n=0
$$
and
$$
\lim_{n\to\infty} nb_n=\lim_{n\to\infty}n\frac{1}{n}=1
$$
If we take instead $c_n=\frac{1}{n^2}$ we have
$$

\lim_{n\to\infty} c_n=0,\qquad
\lim_{n\to\infty} nc_n=\lim_{n\to\infty}\frac{1}{n}=0
$$
With $d_n=\frac{1}{\sqrt{n}}$ we have
$$
\lim_{n\to\infty} d_n=0,\qquad
\lim_{n\to\infty} nd_n=\lim_{n\to\infty}\sqrt{n}=\infty
$$

So no general rule can be given.

You may want to check that
$$
\lim_{n\to\infty} n^a\sin\frac{1}{n^b}=
\begin{cases}
\infty & \text{if $a1 & \text{if $a=b$}\\
0 & \text{if $a>b$}
\end{cases}
$$