×

# 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
3 years, 10 months ago

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 :) · 3 years, 10 months ago

Nice solution · 3 years, 10 months ago

Strangely, this reminds me of Pell's Equations. · 3 years, 10 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 $$\sqrt{3}$$. · 3 years, 10 months ago

I'm assuming you mean $$A, B, n$$ to be positive integers? Otherwise the problem is unsolvable. · 3 years, 10 months ago

It is not necessary ,

Sometimes we even calculate $$\displaystyle \lim_{n \to \infty} f(n!) \dots$$ . · 3 years, 10 months ago