I am trying to recalculate an exam and the solution for my problem is shown in the picture. How ever
$$\lim_{n\to\infty}\left|\dfrac{(n+1)4^{n+1}}{n4^n}\right|=\lim_{n\to\infty}\left|\dfrac{4n+4}{n}\right|=4$$
Could someone please explain to me why the first limit can be reduced to that second one, and then why this is 4?
Thank you!
$$ \frac{(n+1)4^{n+1}}{n4^n}=\frac{(n+1)(4^n)(4)}{n4^n}=\frac{(n+1)4}{n}=\frac{4n+4}{n}. $$
$$ \frac{4n+4}{n}=\frac{4n+4}{n}\cdot \frac{\frac{1}{n}}{\frac{1}{n}}=\frac{4+\frac{4}{n}}{1}=4+\frac{4}{n}. $$
$$ \lim_{n\to \infty}\left[4+\frac{4}{n}\right]=4+\lim_{n\to \infty}\left[\frac{4}{n}\right]=4+4\lim_{n\to \infty}\left[\frac{1}{n}\right]=4+4\cdot 0=4. $$