Is the following language context-free? $L= \{a^nb^m| m\geq2^n\}$
Let $p$ be the integer in Ogden's lemma, and consider the word $a^q b^{2^q} \in L$ for $q$ satisfying $2^q &gt; p$, marking its $b$ part. According to the lemma, we can write $a^q b^{2^q} = uxyzv$ where $xz$ contains between one and $p$ letters $b$. Since $uyv \in L$ and $xz$ contains at least one $b$, it follows that it also contains at least one $a$. For $ux^2yz^2v$ to be in $L$, we need $xz$ to contain at least $2^{q+1}-2^q = 2^q$ letters $b$, contradicting our choice of $2^q &gt; p$.

More generally, this works for any language of the form $\{a^nb^m : m \geq f(n)\}$ for any $f(n) = \omega(n)$, or even whenever $\limsup f(n)/n = \infty$.