# Need help with limits

If $$\left( 1 + \sqrt{3} \right)^n = A + B\sqrt{3}$$, show that the limit of $$\frac{A}{B}$$ as $$n$$ approaches infinity is equal to sqrt(3).

Note by Haris Bin Zahid
4 years, 9 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

It must be $$\sqrt{3}$$ rather. Two ways to solve: $$(1 + \sqrt{3})^n = \displaystyle \sum_{r = 0}^{n} {n \choose r}({\sqrt{3}})^r = A + B\sqrt{3}$$

$$(1 - \sqrt{3})^n = \displaystyle \sum_{r = 0}^{n} {n \choose r}({-\sqrt{3}})^r = A - B\sqrt{3}$$ (Check that $$B$$ comes from odd powers and $$A$$ from even)

Solve to get :

$$A = \frac{(1 + \sqrt{3})^n + (1 - \sqrt{3})^n}{2}$$ ,

$$B = \frac{ (1 + \sqrt{3})^n - (1 - \sqrt{3})^n}{2\sqrt{3}}$$

Now, $$\displaystyle \lim_{n \to \infty} \frac{A}{B}$$

= $$\frac{\frac{(1 + \sqrt{3})^n + (1 - \sqrt{3})^n}{2}}{\frac{ (1 + \sqrt{3})^n - (1 - \sqrt{3})^n}{2\sqrt{3}}}$$

= $$\boxed{\sqrt{3}}$$ . $$\Bigg(As \displaystyle \lim_{n \to \infty} \frac{(1 - \sqrt{3})^n}{(1 + \sqrt{3})^n} = 0 \Bigg)$$

$ALTERNATE:$

$$\displaystyle \lim_{n \to \infty} (1 - \sqrt{3})^n = \displaystyle \lim_{n \to \infty} A - B\sqrt{3} = 0$$ [As $$\displaystyle \lim_{n \to \infty} (1 - \sqrt{3})^n = 0$$]

$$\Rightarrow \displaystyle \lim_{n \to \infty} \frac{A}{B} = \boxed{\sqrt{3}}$$

This was less a limit than a binomial theorem question :)

- 4 years, 9 months ago

Nice solution

- 4 years, 9 months ago

Strangely, this reminds me of Pell's Equations.

- 4 years, 9 months ago

Not surprising. The integer solutions of the equation $x^2 - 2y^2 \; = \; \pm1$ are precisely the integer coefficients $$x + y\sqrt{2} = \pm(1 + \sqrt{2})^n$$ for $$n \in \mathbb{Z}$$ (you get $$+1$$ when $$n$$ is even, and $$-1$$ when $$n$$ is odd). Similar games can be played with $$\sqrt{3}$$.

- 4 years, 9 months ago

I'm assuming you mean $$A, B, n$$ to be positive integers? Otherwise the problem is unsolvable.

- 4 years, 9 months ago

It is not necessary ,

Sometimes we even calculate $$\displaystyle \lim_{n \to \infty} f(n!) \dots$$ .

- 4 years, 9 months ago