Some Important Limits

Appendix B 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 — Figure B.1. A `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 \(c\text{,}\) \(\ds \lim_{x\to \infty} \frac{x^c}{e^x} = 0\text{.}\)

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