Is $(n^5 + n^7)\in \Omega(n^7)$? Shouldn't it be in $\Omega(n^5)$?
I understand Omega to be a "lower bound" on a function. Shouldn't the largest lower bound on the function $n^5 + n^7$ be $n^5$? (Just as the smallest upper bound is $n^7$)
The reason I say that the function is in the Omega class of functions larger than $n^5$ is because of the limit definitions of complexity [Source]: $\require{enclose}$
$$\begin{array}{c|c|c} \text{Big-}\mathcal O\ \text{notation} & \text{Comparison notation} & \text{Limit definition}\\ \hline f \in \mathcal o(g) & f \; {\scriptstyle \enclose{circle}{\kern .07em \bbox[6px] \lt \kern .07em}}\ g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} = 0\\ f \in \mathcal O(g) & f\; {\scriptstyle \enclose{circle}{\kern .07em \bbox[6px]\le \kern .07em}}\ g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} < \infty \\ f \in \Theta(g) & f\; {\scriptstyle \enclose{circle}{\bbox[7px]=}}\ g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} \in \mathbb{R}_{>0} \\ f \in \Omega(g) & f \;{\scriptstyle \enclose{circle}{\kern .07em \bbox[6px]\ge \kern .07em}}\ g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} > 0 \\ f \in \omega(g) & f\; {\scriptstyle \enclose{circle}{\kern .07em \bbox[6px]\ge \kern .07em}}\; g & \lim_{x\rightarrow\infty} \frac{f(x)}{g(x)} = \infty \\ \end{array}$$
Using the limit definition of $\mathcal O$, we correctly identify that $n^5+n^7 \in \mathcal O(n^7)$ and $n^5+n^7 \notin \mathcal O(n^6)$
However, using the limit definition of $\Omega$ tells us that the function is in $\Omega(n^7)$! But isn't the largest lower bound for this function $n^5$?