Limits of Sequences

Denominator's fastest term: $n^2$.

$\dfrac{4 - \frac{1}{n} + \frac{7}{n^2}}{2 + \frac{5}{n^2}}$.

$\to \dfrac{4 - 0 + 0}{2 + 0}$.

$= 2$.

$\sqrt{4n + 1}$ grows like $\sqrt{n}$.

$\dfrac{3 + \frac{2}{\sqrt{n}}}{\sqrt{4 + \frac{1}{n}}}$ (inside the root, divide by $n$).

$\to \dfrac{3 + 0}{\sqrt{4 + 0}}$.

$= \dfrac{3}{2}$.

$5^n$ beats $4^n$.

$\dfrac{1 + \frac{n^3}{5^n}}{3 + \left(\frac{4}{5}\right)^n}$.

$\dfrac{n^3}{5^n} \to 0$ and $\left(\tfrac{4}{5}\right)^n \to 0$, so $\to \dfrac{1}{3}$.

$= \dfrac{1}{3}$.

Denominator's fastest term: $n^2$.

$\dfrac{n + \frac{1}{n^2}}{1 + \frac{2}{n}}$.

The top grows without bound; the bottom $\to 1$.

The sequence diverges to $\infty$.

$-1 \le \cos(3n) \le 1$, so $1 \le 2 + \cos(3n) \le 3$.

Divide by $n^2 > 0$: $\dfrac{1}{n^2} \le a_n \le \dfrac{3}{n^2}$.

$\dfrac{1}{n^2} \to 0$ and $\dfrac{3}{n^2} \to 0$.

By the squeeze theorem, $a_n \to 0$.

$-1 \le (-1)^n \le 1$.

$-\dfrac{n}{n^2 + 4} \le a_n \le \dfrac{n}{n^2 + 4}$.

$\dfrac{n}{n^2 + 4} = \dfrac{1/n}{1 + 4/n^2} \to 0$, and so does its negative.

$a_n \to 0$: it converges.

$\dfrac{n!}{n^n} = \dfrac{1}{n}\cdot\dfrac{2}{n}\cdots\dfrac{n}{n}$, and each factor is between $0$ and $1$.

Keep $\dfrac{1}{n}$, bound the rest by $1$: $0 < \dfrac{n!}{n^n} \le \dfrac{1}{n}$.

$0 \to 0$ and $\dfrac{1}{n} \to 0$.

By the squeeze theorem, $\dfrac{n!}{n^n} \to 0$.

$f(x) = \dfrac{\log x}{x^{1/3}}$.

Already $\dfrac{\infty}{\infty}$.

$\dfrac{1/x}{\tfrac{1}{3}x^{-2/3}} = \dfrac{3}{x^{1/3}} \to 0$.

No log was taken, so $a_n \to 0$.

$f(x) = \left(1 + \dfrac{5}{x}\right)^{2x}$, a $1^\infty$ form.

$\log f(x) = 2x\log\left(1 + \dfrac{5}{x}\right) = \dfrac{2\log\left(1 + \frac{5}{x}\right)}{1/x}$, a $\tfrac{0}{0}$ form.

$\dfrac{2\cdot\dfrac{-5/x^2}{1 + 5/x}}{-1/x^2} = \dfrac{10}{1 + 5/x} \to 10$.

$\log f(x) \to 10$, so $a_n \to e^{10}$.

$f(x) = \left(x^2 + 1\right)^{1/x}$, an $\infty^0$ form.

$\log f(x) = \dfrac{\log(x^2 + 1)}{x}$, an $\tfrac{\infty}{\infty}$ form.

$\dfrac{2x/(x^2 + 1)}{1} = \dfrac{2x}{x^2 + 1} \to 0$.

$\log f(x) \to 0$, so $a_n \to e^0 = 1$.

$f(x) = \dfrac{x^2}{e^x}$.

Already $\dfrac{\infty}{\infty}$.

$\dfrac{2x}{e^x}$, still $\dfrac{\infty}{\infty}$; again: $\dfrac{2}{e^x} \to 0$.

No log was taken, so $a_n \to 0$.

$\dfrac{\left(\sqrt{n^2 + 6n} - n\right)\left(\sqrt{n^2 + 6n} + n\right)}{\sqrt{n^2 + 6n} + n}$.

$\dfrac{n^2 + 6n - n^2}{\sqrt{n^2 + 6n} + n} = \dfrac{6n}{\sqrt{n^2 + 6n} + n}$.

$\dfrac{6}{\sqrt{1 + \frac{6}{n}} + 1}$.

$\to \dfrac{6}{1 + 1} = 3$.

Multiply by $\dfrac{\sqrt{n^2 + 3n} + \sqrt{n^2 - n}}{\sqrt{n^2 + 3n} + \sqrt{n^2 - n}}$.

$\dfrac{(n^2 + 3n) - (n^2 - n)}{\sqrt{n^2 + 3n} + \sqrt{n^2 - n}} = \dfrac{4n}{\sqrt{n^2 + 3n} + \sqrt{n^2 - n}}$.

$\dfrac{4}{\sqrt{1 + \frac{3}{n}} + \sqrt{1 - \frac{1}{n}}}$.

$\to \dfrac{4}{1 + 1} = 2$.

Multiply by $\dfrac{\sqrt{n^2 + n} + 2n}{\sqrt{n^2 + n} + 2n}$.

$\dfrac{n^2 + n - 4n^2}{\sqrt{n^2 + n} + 2n} = \dfrac{-3n^2 + n}{\sqrt{n^2 + n} + 2n}$.

$\dfrac{-3n + 1}{\sqrt{1 + \frac{1}{n}} + 2}$.

The top $\to -\infty$, the bottom $\to 3$: the sequence diverges to $-\infty$.

$L = \dfrac{L + 10}{2}$.

$2L = L + 10$, so $L = 10$.

$a_{n+1} - 10 = \dfrac{a_n + 10}{2} - 10 = \dfrac{a_n - 10}{2}$.

The gap halves each step: $|a_n - 10| = \dfrac{10}{2^{n-1}} \to 0$.

$L = 6 - \dfrac{L}{2}$.

$\dfrac{3L}{2} = 6$, so $L = 4$.

$a_{n+1} - 4 = 2 - \dfrac{a_n}{2} = -\dfrac{a_n - 4}{2}$.

The gap halves and switches sides ($1$, $5.5$, $3.25$, …), so $a_n \to 4$.

$L = 3L - 4$.

$L = 2$.

$a_{n+1} - 2 = 3a_n - 6 = 3(a_n - 2)$.

The gap triples: $a_n - 2 = -3^{n-1}$, so $a_n = 2 - 3^{n-1} \to -\infty$ ($1$, $-1$, $-7$, …).