conjugate trick

Conjugate trick

To simplify expressions of the form $\sqrt{a}-\sqrt{b}$, multiply and divide by the conjugate $\sqrt{a}+\sqrt{b}$:

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

This turns a difference of roots into a rational expression (no root in the numerator).

Useful whenever a limit involves $\sqrt{\cdot }-\sqrt{\cdot }$, which looks like $\infty -\infty$ (indeterminate). After the trick, the expression usually simplifies.

${\lim }_{n\to \infty }(\sqrt{n+1}-\sqrt{n})$

$$\sqrt{n+1}-\sqrt{n}=\frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$$

${\lim }_{n\to \infty }(\sqrt{n^2+2n}-n)$

$$\sqrt{n^2+2n}-n=\frac{n^2+2n-n^2}{\sqrt{n^2+2n}+n}=\frac{2n}{\sqrt{n^2+2n}+n}$$

Divide numerator and denominator by $n$:
Numerator: $\frac{2n}{n}=2$
Denominator: $\frac{\sqrt{n^2+2n}}{n}+\frac{n}{n}=\sqrt{\frac{n^2+2n}{n^2}}+1=\sqrt{1+\frac{2}{n}}+1$

(we pulled $n=\sqrt{n^2}$ inside the root)

$$=\frac{2}{\sqrt{1+2/n}+1}\stackrel{n\to \infty }{\to }\frac{2}{\sqrt{1}+1}=1$$