Find $\lim_{n \to \infty} \frac{a_n}{b_n}$

Question

Suppose $\langle a_n\rangle$ and $\langle b_n\rangle$ are two convergent sequences of real numbers such that $a_n > 0$ and $b_n > 0$ for all n. Suppose $\lim_{n \to \infty} a_n = a$ and $\lim_{n \to \infty} b_n = b$. Let $c_n = \frac{a_n}{b_n}$. Then

1. $\langle c_n\rangle$ converges if $b \gt 0$

2. $\langle c_n\rangle$ converges only if $a = 0$

3. $\langle c_n\rangle$ converges only if $b \gt 0$

4. $\limsup_{x \to \infty}c_n = \infty$ if $b = 0$

My Attempt:

Since $\langle a_n\rangle$ and $\langle b_n\rangle$ are positive terms sequences so $\lim_{n \to \infty}a_n = a \gt 0$ and $\lim_{n \to \infty}b_n = b \gt 0$ Also $\langle a_n\rangle$ and $\langle b_n\rangle$ are convergent. So $a$ and $b$ are finite. Therefore $\lim_{n \to \infty}c_n = \frac{a}{b} \gt 0$ and finite. Hence using above information, we say that option 1 is true and options 2,3 are false. Also we know that if $\lim_{n \to \infty}c_n = L$ then $\limsup_{n \to \infty} = L$ and $\liminf_{n \to \infty} = L$. Hence option 4 is false. I'm right ? If not then answer my question or point out my mistake. Thanks in advance.

Answer 1

Strict inequalities are transformed to loose inequalities by taking the limits.

Therefore $a_n,b_n>0$ implies only $a,b\ge 0$ and zero cannot be excluded.

Let start with the case $b=0$.

• $a_n=\dfrac 1n$ and $b_n=\dfrac 1{n^2}$ then $c_n=n\to\infty$ so $(2)$ is not true.

• $a_n=\dfrac 3n$ and $b_n=\dfrac 1n$ then $c_n=3\to 3$ so $(3)$ and $(4)$ are not true.

It remains the case $b>0$

• In that case $c_n\to \dfrac ab$ is well defined and $(1)$ is true.