Why do I get two different answers while doing this simple limit math?

Question

Problem:

$$\lim_{n\to\infty} \frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}$$

Attempt 1:

$$\lim_{n\to\infty} \frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}$$

$$=\lim_{n\to\infty} \frac{(n+1)(2n+1)}{6n^2}$$

$$=\lim_{n\to\infty} \frac{n+1}{6n}\cdot\frac{2n+1}{n}$$

$$=\lim_{n\to\infty} \left(\frac{1}{6}+\frac{1}{6n}\right)\cdot\left(2+\frac{1}{n}\right)$$

$$=\left(\frac{1}{6}+0\right)\cdot\left(2+0\right)$$

$$=\frac{1}{6}\cdot2$$

$$=\frac{1}{3}$$

Attempt 2:

$$\lim_{n\to\infty} \frac{1}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}$$

$$=\lim_{n\to\infty} \frac{1}{n^3}\cdot\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6}$$

$$=0\cdot\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6}$$

$$=0$$

I know that attempt 1 is the correct attempt; however, I'm unsure as to why attempt 2 is not working. Why is attempt 2 not working?

Answer 1

In attempt 2, your method uses the fact that if you have two convergent sequences $a_n\to a<\infty$, $b_n\to b<\infty$, then the product is also convergent: $a_nb_n\to ab<\infty$. The issue you have when you split up your limit like that is that the individual sequences do not converge. In particular, $\frac{n(n+1)(2n+1)}{6}\to\infty$.

This is the same reason we cannot define a function $f(x)=\frac{x}{x}=1$ and then say: \begin{align*} 1=\lim_{x\to 0}f(x)=\lim_{x\to 0}\frac{x}{x}"="\lim_{x\to 0}x\lim_{x\to 0}\frac{1}{x}=0\times\lim_{x\to 0}\frac{1}{x}=0 \end{align*} A nice rule of thumb is: If you aren't sure if a limit exists, don't write it down.

Answer 2

Another way to look at it is:

$$\lim_{n\to\infty} \frac{1}{n^3}\cdot\lim_{n\to\infty}\frac{n(n+1)(2n+1)}{6}$$

$$=0\cdot \infty $$

$$=undefined$$

$0 \cdot X $ is not $0$, if $X$ is tends to $ \infty $

You have basically converted 1 undefined form to another i.e. $\frac{\infty}{\infty}$ to $0\cdot\infty$