conjugate trick

Conjugate trick

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

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

It also works for expressions of the form $\sqrt{a}-b$, or $a-\sqrt{b}$, because $a=\sqrt{a^2}$.
The underlying identity is $(a-b)(a+b)=a^2-b^2$

Useful whenever a limit involves $\sqrt{\cdot }-\sqrt{\cdot }$, which looks like $\infty -\infty$ (indeterminate).
It eliminates the roots from the numerator. The new numerator $a-b$ is typically a polynomial (or at least root-free), which is easier to analyze.

How can the difference containing a squqare root be simplified into a fraction with a rational denominator?

By multiplying and dividing by the conjugate $\sqrt{a}+\sqrt{b}$:

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

Why is the conjugate trick useful for limits involving $\sqrt{\cdot }-\sqrt{\cdot }$?

The expression $\sqrt{a}-\sqrt{b}$ can be indeterminate (like $\infty -\infty$) when $a,b\to \infty$. The conjugate trick eliminates the roots from the numerator, yielding a rational expression $\frac{a-b}{\sqrt{a}+\sqrt{b}}$, which is easier to analyze for limits.

Apply the conjugate trick to compute ${\lim }_{n\to \infty }(\sqrt{n+1}-\sqrt{n})$.

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

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

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

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)

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