Express each limit in terms of the number $e$:

Solution (a) This limit resembles $\lim\limits_{h\rightarrow 0}(1+h) ^{1/h}$, and with some manipulation, it can be expressed in terms of $\lim\limits_{h\rightarrow 0}(1+h) ^{1/h}$. $$( 1+2h) ^{1/h}= [ ( 1+2h) ^{1/( 2h) } ] ^{2}$$

Now let $k=2h,$ and note that $h\rightarrow 0$ is equivalent to $2h=k\rightarrow 0$. So, $$\begin{eqnarray*} && \lim\limits_{h\rightarrow 0}( 1+2h) ^{1/h}=\lim\limits_{k\rightarrow 0}\left[ \left( 1+k\right) ^{1/k}\right] ^{2}=\left[ \lim\limits_{k\rightarrow 0}\left( 1+k\right) ^{1/k}\right] ^{2} \underset{\underset{\color{#0066A7}{\hbox{\((2)\)}}}{\color{#0066A7}{\uparrow}}}{=} e^{2} \\[-10pt] && \end{eqnarray*}$$

(b) This limit resembles $\lim\limits_{n\rightarrow \infty} \left( 1+\dfrac{1}{n}\right) ^{n}$. We rewrite it as follows: $$\left( 1+\dfrac{3}{n}\right) ^{2n}=\left[ \left( 1+\dfrac{3}{n}\right) ^{2n} \right] ^{3/3}=\left[ \left( 1+\dfrac{3}{n}\right) ^{n/3}\right] ^{6}$$

Let $k=\dfrac{n}{3}$. Since $n\rightarrow \infty$ is equivalent to $\dfrac{n }{3}=k\rightarrow \infty \,$, we find that $$\begin{eqnarray*} && \lim\limits_{n\rightarrow \infty }\left( 1+\dfrac{3}{n}\right) ^{2n}=\lim\limits_{k\rightarrow \infty }\left[ \left( 1+\dfrac{1}{k}\right) ^{k}\right] ^{6}=\left[ \lim\limits_{k\rightarrow \infty }\left( 1+\dfrac{1}{ k}\right) ^{k}\right] ^{6} \underset{\underset{\color{#0066A7}{\hbox{\((2)\)}}}{\color{#0066A7}{\uparrow}}}{=} e^{6} \end{eqnarray*}$$