root

$n$-th root

For $x\ge 0$ and $n\in \mathbb{N}$:

$$\sqrt[n]{x}=x^{1/n}=y⟺y^n=x\text{ and }y\ge 0$$

$\sqrt{x^2}=|x|$

Not $x$, because $x$ might be negative. E.g. $\sqrt{(-3{)}^2}=\sqrt{9}=3=|-3|$.

Properties

For $a,b\ge 0$:

$$\sqrt[n]{a\cdot b}=\sqrt[n]{a}\cdot \sqrt[n]{b},\sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}(b\ne 0)$$

In particular, dividing “through” a root:

$$\frac{\sqrt{x}}{a}=\frac{\sqrt{x}}{\sqrt{a^2}}=\sqrt{\frac{x}{a^2}}(a>0)$$

This is useful for simplifying limits: e.g. $\frac{\sqrt{n^2+2n}}{n}=\sqrt{\frac{n^2+2n}{n^2}}=\sqrt{1+\frac{2}{n}}$

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$$\underset{n\to \infty }{\lim }\sqrt[n]{x}=1\text{for all }x>0$$ $$\underset{n\to \infty }{\lim }\sqrt[n]{g(n)}=1\text{for all }g(n)\dots$$

that grow subexponentially.

How can we simplify $\frac{\sqrt{n^2+2n}}{n}$ to compute ${\lim }_{n\to \infty }\frac{\sqrt{n^2+2n}}{n}$?

Put the expression under a single root by pulling the denominator inside:

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

Simplify $2^{5/2}$

$$2^{5/2}=\sqrt{2^5}=\sqrt{32}=\sqrt{16\cdot 2}=4\sqrt{2}$$

Alternatively

$$2^{5/2}=2^{2+1/2}=2^2\cdot 2^{1/2}=4\cdot \sqrt{2}$$