Useful Asymptotic Calculations

$\sqrt{(n-1)\log n}=(1+o(1))\sqrt{n\log n}$

We first start with the following inequality

$$\sqrt{(n-1)\log n}=\sqrt{n\log n-\log n}=\sqrt{n\log n\cdot \left(1+\frac{1}{n}\right)}=\sqrt{n\log n}\cdot \sqrt{1+\frac{1}{n}}$$

Let $x=1/n$, and for large values of $n\to \infty$ we have that $x\to 0^{+}$. For $x=0$, the function $f(x)=\sqrt{1-x}$ has Taylor expansion

$$\begin{array}{rl}\sqrt{1-x}& =f(0)+f^{\prime }(0)(x-0)+o(x)\\ & =1-\frac{x}{2}+o(x)\end{array}$$

Remind

Remind

$$\frac{d}{dx}\sqrt{1-x}=\frac{1}{2\sqrt{1-x}}\cdot \frac{d}{dx}(1-x)=-\frac{1}{2\sqrt{1-x}}$$

Since $x=1/n$, we have

$$\sqrt{1-\frac{1}{n}}=1-\frac{1}{2n}+o\left(\frac{1}{n}\right)$$

and thus (for $n\to \infty$)

$$\sqrt{(n-1)\log n}=\sqrt{n\log n}\cdot \left(1-\frac{1}{2n}+o\left(\frac{1}{n}\right)\right)=(1+o(1))\sqrt{n\log n}$$

since $-1/2n+o(1/n)=o(1/n)$.