Appendix C. Some Important Limits
The limit \(\ds \lim_{h \to 0} \frac{\sin(h)}{h}\) is \(1\text{.}\) To see this first observe that the funtion \(\sin(h)/h\) is even. So we just need to argue the right limit is \(1\text{.}\) To that end consider the areas of the two right triangles and the sector of the unit circle in between.
Unit circles and two triangles
diagram for computing the limit \(\ds \lim_{h\to 0} \sin(h)/h\)We have
\begin{equation*}
\frac{1}{2}\sin(h)\cos(h) \le \frac{1}{2}h \le
\frac{1}{2}\tan(h).
\end{equation*}
Canceling the factor \(1/2\) and rearranging the terms, we get
\begin{equation*}
\cos(h) \le \frac{\sin{h}}{h}
\le \frac{1}{\cos(h)}.
\end{equation*}
Since \(\cos(h) \to 1\) as \(h\to 0\text{,}\) we conclude that \(\ds
\lim_{h \to 0} \frac{\sin(h)}{h} = 1\text{.}\)
A consequence of the limit above is that \(\ds \lim_{h \to 0}
\frac{1-\cos(h)}{h} = 0\text{.}\) This is because
\begin{equation*}
\frac{1-\cos(h)}{h} = \frac{1}{h}\frac{\sin^2(h)}{1+\cos(h)} =
\sin(h)\frac{\sin(h)}{h}\frac{1}{1+\cos(h)}
\end{equation*}
and the right-hand side of the equation about tends to \((0)(1)(1/2) = 0\text{.}\)
The limit \(\ds \lim_{x \to \infty} \frac{\ln(x)}{x} =0\text{.}\) We show this without using the L’Hopital rule. Since \(1/x\) is a decreasing function, for any \(x \ge 1\text{,}\)
\begin{equation*}
\ln(x) = \int_1^x \frac{1}{t} dt \le \int_1^x dt = x-1 \le x.
\end{equation*}
Note that for \(x \ge 1\text{,}\) \(\sqrt{x} \ge 1\) as well. Therefore,
\begin{equation*}
\frac{1}{2}\ln(x) = \ln(\sqrt{x}) \le \sqrt{x}.
\end{equation*}
And we conclude that
\begin{equation*}
0 \le \frac{\ln(x)}{x} \le \frac{2}{\sqrt{x}}.
\end{equation*}
From this, according to the squeeze lemma, it follows that \(\ln(x)/x
\to 0\) as \(x \to \infty\text{.}\)
Furthermore, for any constant \(c\text{,}\) as \(x \to \infty\)
\begin{equation*}
\frac{\ln(x^c)}{x} = \frac{c\ln(x)}{x} \to 0.
\end{equation*}
In particular, for all \(x\) sufficiently large, \(\ln(x^c)/x \lt 1/2\) and so \(\ln(x^c) \lt x/2\text{.}\) Applying the exponential function on both side, we get \(x^c \lt e^{x/2}\text{.}\) Therefore,
\begin{equation*}
0 \le \frac{x^c}{e^x} \lt {e^{-x/2}}
\end{equation*}
Letting \(x \to \infty\text{,}\) we see that for any \(c\text{,}\) \(\ds \lim_{x\to
\infty} \frac{x^c}{e^x} = 0\text{.}\)
