<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] Hint
For large values of $k$, the following Taylor expansion can be built $$\frac{2+\sqrt{3+9k^2}}{3k-1}=1+\frac{1}{k}+\frac{1}{2 k^2}+\frac{1}{6 k^3}+\frac{1}{24 k^4}+\frac{1}{72
   k^5}+O\left(\left(\frac{1}{k}\right)^6\right)$$ So $$\log\Big(\frac{2+\sqrt{3+9k^2}}{3k-1}\Big)=\frac{1}{k}+\frac{1}{180 k^5}+O\left(\left(\frac{1}{k}\right)^6\right)$$
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