Question

Evaluate the limit of sequence:

Answer 1

Rationalize both the numerator and denominator. Given

$\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d}$

we can rationalize it by introducing conjugates of the numerator and denominator:

$\dfrac{\sqrt a-\sqrt b}{\sqrt c-\sqrt d} \cdot \dfrac{\sqrt a+\sqrt b}{\sqrt a+\sqrt b} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt c+\sqrt d} \\\\ = \dfrac{\left(\sqrt a\right)^2 - \left(\sqrt b\right)^2}{\left(\sqrt c\right)^2-\left(\sqrt d\right)^2} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b} \\\\ = \dfrac{a-b}{c-d} \cdot \dfrac{\sqrt c+\sqrt d}{\sqrt a + \sqrt b}$

Then the limit is equivalent to

$\displaystyle \lim_{n\to\infty} \frac{(n+3)-n}{(n+1)-n} \cdot \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n} = 3 \lim_{n\to\infty} \dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n}$

For the remaining expression, divide through uniformly by $\sqrt n$:

$\dfrac{\sqrt{n+1}+\sqrt n}{\sqrt{n+3}+\sqrt n} = \dfrac{\sqrt{1+\frac1n} + 1}{\sqrt{1+\frac3n}+1}$

As n goes to infinity, the remaining terms containing n converge to 0, leaving

$\dfrac{\sqrt{1}+1}{\sqrt1+1} = \dfrac22 = 1$

making the overall limit 3.