Q: What is the most simplified answer in Θ notation for $\ f(n) = n^\frac{(3loglogn)}{logn + 5}$.
I know to drop the constants to make it $\ ( n^\frac{(loglogn)}{logn})$, and that the exponential $\lim_{x\to ∞}$ $\frac{(loglogn)}{logn} = 0$. Would that mean that $\ f(n) = Θ(1) $ since $\ c^0 = 1 $?
On
On the one hand $n^{\frac{ 3\log \log n}{\log n+5}} \le n^{\frac{3\log \log n}{\log n}} = \left(n^{\frac{1}{\log n}}\right)^{3 \log \log n} = a^{3 \log \log n} = \log^3_a n$.
[The constant $a$ is s.t. $\log x = \log_a x$.]
A. So $n^{\frac{ 3\log \log n}{\log n+5}} \le \log^3_a n$.
On the other hand:
$n^{\frac{ 3\log \log n}{\log n+5}} = a^{3 \log \log n\frac{\log n}{\log n+5}} \ge a^{3 \log \log n\frac{\log n-5}{\log n}}$ $=a^{3 \log \log n} \times a^{-5 \times \frac{3 \log \log n}{\log n}}$ $= a^{3 \log \log n} \times 2^{-5 \times \frac{3 \log_a \log_a n}{\log_2 n}}$ $\ge a^{3 \log \log n} \times 2^{-5}$ $=\frac{\log^3_a n}{32}$.
B. So $n^{\frac{ 3\log \log n}{\log n+5}} \ge \frac{\log^3 n}{32}$.
Thus putting A and B together yields $n^{\frac{ 3\log \log n}{\log n+5}} = \theta(\log^3_a n)$.
No what you've said doesn't follow. Otherwise you'd be able to do things like: $$e = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n {"="} \lim_{m\to\infty}\left(\lim_{n\to\infty}1+\frac{1}{n}\right)^m = \lim_{m\to\infty}1^m = 1$$
Also the $3$ in the numerator is not a constant multiple of (or addition to) the function, so you may not just get rid of a constant in the exponent. So $$n^\frac{3\log \log n}{\log n} \not\in \Theta(n^\frac{\log \log n}{\log n})$$
Instead, just use your basic log properties: $$f(n) = n^{\frac{3\log\log(n)}{\log(n)+5}} = \exp\left(3 \log\log(n) \frac{\log(n)}{\log(n)+5}\right) $$ Note that the right-most fraction approach $1$, whereas $\log\log(n)\to\infty$, so we think the behavior is like $\exp\left(3\log\log(n)\right)$. To confirm: $$\lim_{n\to\infty} \frac{f(n)}{\exp\left(3\log\log(n)\right)} = \lim_{n\to\infty} \exp\left(-15 \frac{\log\log(n)}{\log(n)+5}\right) = \exp(0) =1$$ Hence $f(n) \in \Theta(e^{3\log\log(n)})$, or more neatly: $$\exp\left(3\log\log(n)\right) = [\exp\left(\log(\log(n))\right)]^3 = \log(n)^3$$ so $f(n) \in \Theta(\log(n)^3)$.