Limits of Sequences practice questions — AP Calculus BC

With $a = 2$: $\!\left(1 + \dfrac{2}{n}\right)^n \to e^2$.

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

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

By the squeeze theorem, $-\dfrac{1}{n} \le \dfrac{\sin n}{n} \le \dfrac{1}{n}$ 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 $e$: $\!\left(1 + \dfrac{1}{n}\right)^n \to e$.

$\dfrac{n + 1}{n^2 + 3} = \dfrac{1/n + 1/n^2}{1 + 3/n^2} \to \dfrac{0}{1} = 0$.

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

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

$\!\left(\dfrac{1}{2}\right)^n \to 0$ since $|q| = \dfrac{1}{2} < 1$.