I'm trying to find a way to simplify these two functions: What I did with the first one is factor out $n^7$ and after that I realised that as n approached infinity, the function goes to 0. The problem is, that we're given a multiple choice of upper bounds and the answer is $n^{-1}$ but I don't understand why. I thought it should be $n^{-2}$.
$$\sqrt{n} \cdot \frac{10 + n^{5} + log_2n}{n^7 + n^2 + 100} $$
The second function is this: $$n^{0.99} \cdot (n^{-3} + log_2n^n)$$
Here I used the logarithm power rule to bring the power of n out of the log to get $n\cdot log_2n$ but other than that I have no idea why the answer is $n^2log_n2$
In the first expression using the dominating terms in numerator and denominator leads to $$ \frac{n^{1/2}n^5}{n^7} \approx n^{-3/2} . $$ If you want an integer exponent, that is smaller than $1/n$. The inequality for $1/n^2$ goes the other way.
In the second expression the first term in parentheses goes to $0$ while the second grows. Asymptotically you have $$ n^{0.99}n \log_2(n) \approx n^{1.99}\log_2(n) . $$ If you want an integer exponent, that's slower growth than $n^2\log_2(n)$.