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Solution:

We start by showing that
\begin{equation}\label{30-5-1}
\lim_{x\to 0}\left(1+x\right)^{1/x}=e.
\end{equation} Indeed, note that
$$
\left(1+x\right)^{1/x}=e^{ \ln\left(1+x\right)^{1/x} }.
$$ By L’Hospital’s Rule, we have
\begin{align*}
\lim_{x\to 0}\ln\left(1+x\right)^{1/x}
=&\ \lim_{x\to 0}\frac{\ln(1+x)}{x}\\
=&\ \lim_{x\to 0}\frac{\frac{1}{1+x}}{1}\\
=&\ 1.
\end{align*} Therefore, by Theorem 20.5, we have
$$
\lim_{x\to 0}\left(1+x\right)^{1/x}=e^1=e.
$$

Apply Exercise 30.4, we also have
\begin{equation}\label{30-5-2}
\lim_{y\to \infty}\left(1+\frac{1}{y}\right)^y=e.
\end{equation}

Part a

Let $u=2x$, then $\dfrac{1}{x}=\dfrac{2}{u}$. We have
\begin{align*}
\lim_{x\to 0}(1+2x)^{1/x}
=&\ \lim_{x\to 0}(1+2x)^{1/x}\\
=&\ \lim_{u\to 0}(1+u)^{2/u}\\
=&\ \lim_{u\to 0}\Big((1+u)^{1/u}\Big)^2\\
=&\ \Big(\lim_{u\to 0}(1+u)^{1/u}\Big)^2\\
\text{By \eqref{30-5-1} }\ =&\ e^2.
\end{align*}

Part b

Let $u=y/2$, then $\dfrac{2}{y}=\dfrac{1}{u}$ and $y=2u$. We have
\begin{align*}
\lim_{y\to \infty}\left(1+\frac{2}{y}\right)^{y}
=&\ \lim_{u\to \infty}\left(1+\frac{1}{u}\right)^{2u}\\
=&\ \lim_{u\to \infty}\left(\Big(1+\frac{1}{u}\Big)^{u}\right)^2\\
=&\ \left(\lim_{u\to \infty}\Big(1+\frac{1}{u}\Big)^{u}\right)^2\\
\text{By \eqref{30-5-2} }\ =&\ e^2.
\end{align*}

Part c

Note that
$$
\ln (e^x+x)^{1/x}=\frac{\ln(e^x+x)}{x}.
$$ By L’Hospital’s Rule, we have
\begin{align*}
\lim_{x\to \infty}\ln (e^x+x)^{1/x}
=&\ \lim_{x\to \infty}\frac{\ln(e^x+x)}{x}\\
=&\ \lim_{x\to \infty}\frac{\frac{e^x+1}{e^x+x}}{1}\\
=&\ \lim_{x\to \infty}\frac{e^x+1}{e^x+x}\\
=&\ \lim_{x\to \infty}\frac{1+\frac{1}{e^x}}{1+\frac{x}{e^x}}\\
=&\ \frac{1+0}{1+0}=1.
\end{align*} Here we used $\lim_{x\to\infty}\dfrac{1}{e^x}=0$ and (using L’Hospital’s Rule below)
$$
\lim_{x\to \infty}\frac{x}{e^x}=\lim_{x\to \infty}\frac{1}{e^x}=0.
$$ Therefore, we conclude using Theorem 20.5 that
$$
\lim_{x\to\infty}(e^x+x)^{1/x}=e^1=e.
$$