Expressing a Limit in Terms of \(e\)

Express each limit in terms of the number \(e\):

1. \(\lim\limits_{h\rightarrow 0}( 1+2h) ^{1/h}\)
2. \(\lim\limits_{n\rightarrow \infty}\left( 1+\dfrac{3}{n}\right) ^{2n}\)

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*} \]