### Limits part 1: Infinite sequences: Techniques

### Conjugate multiplication

Examine the sequence \(a_n = \sqrt{n+1} - \sqrt{n}\) using the animation below.

It is true that \(a_{100} \approx 0.050\) and a calculation shows that \(a_{1000} \approx 0.016\). To prove that \[\lim_{n\to\infty} \left( \sqrt{n+1} - \sqrt{n}\right) = 0\] we cannot use the common rules for limits, because \(\infty - \infty\) is an indeterminate form.

One technique you can use here is multiplication by a conjugate. This technique is based on the fact that for all positive numbers \(a\) and \(b\) we have \[\sqrt a - \sqrt b = \frac{\left(\sqrt a - \sqrt b\right)\left( \sqrt a + \sqrt b \right)}{\sqrt a + \sqrt b} = \frac{a-b}{\sqrt a + \sqrt b }\text. \] Sometimes after manipulating \(a_n\) this way it is possible to apply limit laws to the result. For the \(a_n\) from the example we get \[ \sqrt{n+1} - \sqrt{n} = \frac{(n+1)-n}{ \sqrt{n+1} + \sqrt{n}} = \frac{1}{\sqrt{n+1} + \sqrt{n}}\text.\] With the sum rule for limits the denominator goes to \(\infty\), so this fraction converges to \(0\) .