Limits of Sequences

$2^n$ grows without bound: $\lim 2^n=\mathrm{\infty }$.

$\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{1}{2}}\right)}^n\to 0$ since $\mathrm{\mid }q\mathrm{\mid }={\displaystyle \frac{1}{2}}<1$.

As $n\to \mathrm{\infty }$, ${\displaystyle \frac{1}{n}}\to 0$.

Since $(u_n)$ converges to some limit $\mathrm{\ell }$, taking the limit on both sides of the recurrence gives the fixed-point equation: $$\mathrm{\ell }=\frac{1}{2}\mathrm{\ell }+3$$ $$\frac{1}{2}\mathrm{\ell }=3$$ $$\mathrm{\ell }=6$$

This is the standard definition of $e$: $\text{\hspace{0.17em}⁣}{(1+{\displaystyle \frac{1}{n}})}^n\to e$.

The denominator has higher degree, so the ratio tends to $0$.

${\displaystyle \frac{n+1}{n^2+3}}={\displaystyle \frac{1\mathrm{/}n+1\mathrm{/}{n}^2}{1+3\mathrm{/}{n}^2}}\to {\displaystyle \frac{0}{1}}=0$.

${\displaystyle \frac{3n^2+5}{2n^2-1}}={\displaystyle \frac{3+5\mathrm{/}{n}^2}{2-1\mathrm{/}{n}^2}}\to {\displaystyle \frac{3}{2}}$.

By the squeeze theorem, $-{\displaystyle \frac{1}{n}}\le {\displaystyle \frac{\sin n}{n}}\le {\displaystyle \frac{1}{n}}$ and both bounds tend to $0$, so the limit is $0$.

With $a=2$: $\text{\hspace{0.17em}⁣}{(1+{\displaystyle \frac{2}{n}})}^n\to e^2$.