Evaluating e using limits What algebraic operations can I use on the $RHS$ to show $RHS = LHS$
$$e=\lim_{k\to\infty}\left(\frac{2+\sqrt{3+9k^2}}{3k-1}\right)^k$$
 A: 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
$$\lim_{k \to \infty} \frac{ \frac{3k-1}{2+\sqrt{3+9k^2}} \frac{d}{dk} \left[ \frac{2+\sqrt{3+9k^2}}{3k-1} \right] }{-\frac{1}{k^2}}=\lim_{k \to \infty} -\frac{3 k^2 \left(\frac{2 k}{\sqrt{k^2+\frac{1}{3}}}+1\right)}{1-9 k^2}=1.$$
A: Note the general fact if $a_k \rightarrow a$ then
$$\left(1+\frac{a_k}{k}\right)^k \rightarrow e^a$$ as $k\rightarrow \infty$
So set $$1+\frac{a_k}{k}=\frac{2+\sqrt{3+9k^2}}{3k-1}$$
to get 
$$a_k=\frac{k}{3k-1} (3+\sqrt{3+9k^2}-3k)$$
Now $$\sqrt{3+9k^2}-3k=\frac{3}{\sqrt{3+9k^2}+3k}\rightarrow 0$$
Therefore $a_k\rightarrow 1$ and we have the limit.
A: 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 $$\log\Big(\frac{2+\sqrt{3+9k^2}}{3k-1}\Big)=\frac{1}{k}+\frac{1}{180 k^5}+O\left(\left(\frac{1}{k}\right)^6\right)$$
