Suppose $a,b>0$ and $\frac{a+b}{2}<1$
Why is the following relation true as $n\rightarrow\infty $ $$n(1-\frac{a\log n}{n})^{n/2-1}(1-\frac{b\log n}{n})^{n/2}\sim n^{1-\frac{a+b}{2}}$$
The $1$ in the exponent on the right hand side value is obviously clear... So Why does $(1-\frac{a\log n}{n})^{n/2}(1-\frac{b\log n}{n})^{n/2}$ reduce to $n^{\frac{a+b}{2}}$?
Hint:
Use $\left ( 1+\frac{x}{n} \right)^n \sim e^x$.
I hope this helps ^_^
Consider $$y=\left(1-a\frac{ \log (n)}{n}\right)^{\frac{n}{2}-1}$$ $$\log(y)=\left(\frac{n}{2}-1\right)\log\left(1-a\frac{ \log (n)}{n}\right)\sim -\left(\frac{n}{2}-1\right)a\frac{ \log (n)}{n}$$ So, taking the logarithm of the rhs of you expression $A$ we have $$\log(A)\sim \log(n)-\left(\frac{n}{2}-1\right)a\frac{ \log (n)}{n}-\frac n 2 b \frac{ \log (n)}{n}$$ Expanding and grouping $$\log(A) \sim \left(1-\frac{a+b}{2} \right) \log (n)+a\frac{ \log (n)}{n}\sim \left(1-\frac{a+b}{2} \right) \log (n)$$ $$A=e^{\log(A)} \sim n^{1-\frac{a+b}{2}}$$
