Limits of Sequences practice questions — AP Calculus BC

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

Factor the $n$-th term to expose the first term and the ratio: $$\frac{3^{n+1}}{4^{n-1}}=9{\left(\frac{3}{4}\right)}^{n-1}$$

So the first term is $a=9$ (the $n=1$ value) and the common ratio is $$r=\frac{3}{4}$$

Since $\mathrm{\mid }r\mathrm{\mid }=\frac{3}{4}<1$, the series converges to $$S=\frac{a}{1-r}=\frac{9}{1-\frac{3}{4}}=\frac{9}{\frac{1}{4}}=36$$

$\sqrt{n+1}-\sqrt{n}={\displaystyle \frac{1}{\sqrt{n+1}+\sqrt{n}}}\to 0$.

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 $0$.

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

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

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

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

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

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