Unpacking the definition of negligible & non-negligible
As @SEJPEM mentioned in the comment: intuitively, for every positive
polynomial $p(n)$ ($n\in\mathbb{N}$) that you take, consider the function $q(n)=\frac{1}{p(n)}$, then we say that the function $\epsilon(n)$ is negligible if from some point $N\in\mathbb{N}$ it follows that $\epsilon(n)&lt;q(n)$. This means that $\epsilon(n)$ reaches 0 very fast because it doesn't matter whether you take $p_1(n)=n$ or $p_2(n)=n^{100000}$, by the definition there would be $N_1$ from which $\epsilon(n)&lt;\frac{1}{n}$ and $N_2$ from which $\epsilon(n)&lt;\frac{1}{n^{100000}}$.

Now, to interpret the definition that you copied: $\lambda_d\in\mathbb{N}$ refers to $N$ in my explanation above, then the definition of negligible function requires that for every $\lambda\in\mathbb{N}$ such that $\lambda&gt;\lambda_d$ (in my explanation above: for all $n$ that is bigger than this $N$) it follows that $\epsilon(\lambda)&lt;q_d(\lambda)=\frac{1}{p_d(\lambda)}=\lambda^d$.

In order to show that $\epsilon(\lambda)=\frac{1}{2^\lambda}$ is negligible you need to show that for every polynomial $p(\lambda)$ it follows that $\epsilon(\lambda)=\frac{1}{2^\lambda}&lt;q(\lambda)=\frac{1}{p(\lambda)}$ from some point $N\in\mathbb{N}$ (this point could be different for different $p(\lambda)$-s. Now take some arbitrary $p(\lambda)$, we can discover this $N$ from which $\epsilon(\lambda)&lt;\frac{1}{p(\lambda)}$ as follows: let $p(\lambda)=\lambda^k$ then we want to find $\lambda$ such that $\frac{1}{2^\lambda}&lt;\frac{1}{\lambda^k}$, thus, we want to find $\lambda$ such that $\lambda^k&lt;2^\lambda$, or in other words, we want to prove that from some $\lambda$ the term $2^\lambda$ is bigger than $\lambda^k$, and this can be proven using the definition of a limit (see this proof).

Now take $\epsilon(\lambda)=\frac{1}{\lambda^{1000}}$, in order to show that it is not negligible you need to show that there exists some polynomial $p(\lambda)$ that violates the definition. Take the polynomial $p(\lambda)=\lambda^{2000}$, there exist no $\lambda_{2000}$ (or $N$) from which $\epsilon(\lambda)=\frac{1}{\lambda^{1000}} &lt; q(\lambda)=\frac{1}{p(\lambda)}=\frac{1}{\lambda^{2000}}$ (i.e. for every $\lambda\in\mathbb{N}$ that you take, it follows that $\epsilon(\lambda)&gt;\frac{1}{p(\lambda)}$.

I hope that it helps.