Question

If a>0 and b >0 Show that

lim n->infinity squareroot((n+a)(n+b)) -n = (a+b)/2

Answer 1

Answer on Question 37681 – Math - Real Analysis

If $a > 0$ and $b > 0$ show that

$\lim_{n \to \infty} n - \frac{1}{n} = 0$.

Solution

1. The sequence $1/n$ converges to 0 because for given $\varepsilon > 0$ we can choose $N$ such that $N > 1/\varepsilon$. Then for all $n > N$, $|1/n| < \varepsilon$, so

$$\lim_{n \to \infty} \frac{1}{n} = 0.$$

The sequence $\frac{1}{\sqrt{n}}$ also converges to 0 because for given $\varepsilon > 0$ we can choose $N$ such that $N > 1/\varepsilon^2$. Then for all $n > N$, $\left|\frac{1}{\sqrt{n}}\right| < \varepsilon$, so

$$\lim_{n \to \infty} \frac{1}{\sqrt{n}} = 0.$$

2. If $a > 0$ and $b > 0$ then we have two inequalities

$$\sqrt{1 + \frac{a}{n}} \leq 1 + \frac{\sqrt{a}}{\sqrt{n}} \quad \text{and} \quad \sqrt{1 + \frac{b}{n}} \leq 1 + \frac{\sqrt{b}}{\sqrt{n}}$$

Hence,

$$1 \leq \sqrt{\left(1 + \frac{a}{n}\right)\left(1 + \frac{b}{n}\right)} \leq \left(1 + \frac{\sqrt{a}}{\sqrt{n}}\right) \left(1 + \frac{\sqrt{b}}{\sqrt{n}}\right).$$

Using (2) and the limit laws for sequences we conclude that

$$\lim_{n \to \infty} \left(1 + \frac{\sqrt{a}}{\sqrt{n}}\right) = \lim_{n \to \infty} \left(1 + \frac{\sqrt{b}}{\sqrt{n}}\right) = 1.$$

Hence, by the Squeeze Theorem we can infer

$$\lim_{n \to \infty} \sqrt{\left(1 + \frac{a}{n}\right)\left(1 + \frac{b}{n}\right)} = 1.$$

3. We rationalize the numerator, multiplying by the sum of square roots, and introducing this sum as a denominator, giving

$$\sqrt{(n + a)(n + b)} - n = \frac{(a + b)n + ab}{\sqrt{(n + a)(n + b)} + n}.$$

This expression is an "$\frac{\infty}{\infty}$" type, and we divide the numerator and denominator by $n$ to obtain

$$\sqrt{(n + a)(n + b)} - n = \frac{(a + b) + \frac{ab}{n}}{\sqrt{\left(1 + \frac{a}{n}\right)\left(1 + \frac{b}{n}\right)} + 1} \rightarrow \frac{a + b}{2} \quad \text{as } n \to \infty,$$

since using (1) and (3) we can conclude by the limit laws for sequences that the numerator tends to $a + b$ and the denominator tends to 2 as $n \to \infty$.