<s>[INST] Evaluating e using limits What algebraic operations can I use on the $RHS$ to show $RHS = LHS$
$$e=\lim_{k\to\infty}\left(\frac{2+\sqrt{3+9k^2}}{3k-1}\right)^k$$
 [/INST] Since$$\left( \frac{2+\sqrt{3+9k^2}}{3k-1} \right)^k=e^{k \log \left( \frac{2+\sqrt{3+9k^2}}{3k-1} \right)} $$
it's sufficient to show that
$$\lim_{k \to \infty} k \log \left( \frac{2+\sqrt{3+9k^2}}{3k-1} \right)=1 .$$
We can rewrite this expression as
$$\frac{\log \left(\frac{2+\sqrt{3+9k^2}}{3k-1} \right)}{\frac{1}{k}}, $$
which approaches the indeterminate form $0/0$ as $k \to \infty$. We thus apply L'hôpital's rule, and obtain the equivalent limit
$$\lim_{k \to \infty} \frac{ \frac{3k-1}{2+\sqrt{3+9k^2}} \frac{d}{dk} \left[ \frac{2+\sqrt{3+9k^2}}{3k-1} \right] }{-\frac{1}{k^2}}=\lim_{k \to \infty} -\frac{3 k^2 \left(\frac{2 k}{\sqrt{k^2+\frac{1}{3}}}+1\right)}{1-9 k^2}=1.$$
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