<s>[INST] Find $\lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}$ Find $\displaystyle \lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}$
My attempt:
\begin{align*}
A&=\lim_{k\to 2}\frac{2^{2^k}-2^{k^2}}{2^k-k^2}\\
\ln A&=\lim_{k\to 2}\ln\frac{2^{2^k}-2^{k^2}}{2^k-k^2}\\
&=\lim_{k\to 2}(\ln(2^{2^k}-2^{k^2})-\ln(2^k-k^2))
\end{align*}
I stuck at $\ln(0)$.
 [/INST] Applying L'Hopital's rule you get
$$
\lim_{x\to 2}\dfrac{2^{x+2^x} \log ^2 2-2^{x^2+1} x \log (2)}{2^x \log 2-2 x} = 16 \log 2.
$$
</s>