Miscellaneous Proofs

Tags

Cegep1

Mathematics

Word count

658 words

Reading time

5 minutes

$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$

Proof 1

where $x\in (0,\frac{\pi }{2})$ is the angle between the x-axis and the line $\overline{OC}$.
This creates the following inequality:

$$A_{\mathrm{△}DOB}\le A_{\text{sector }DOB}\le A_{\mathrm{△}COB}$$

$A_{\mathrm{△}DOB}$

The base, $\overline{OB}$, and the hypotenuse, $\overline{DO}$, are radii of the unit circle, so they have length 1.

For the height,

$$\overline{DA}=\sin x\cdot \overline{DO}=\sin x\cdot 1=\sin x$$

Then,

$$A_{\mathrm{△}DOB}=\frac{bh}{2}=\frac{\sin x}{2}$$

$A_{\text{sector }DOB}$

$$A_{\text{sector }DOB}=\frac{1}{2}{r}^2\theta =\frac{x}{2}$$

$A_{\mathrm{△}COB}$

$$\overline{CB}=\tan x\cdot \overline{OB}=\tan x\cdot 1=\tan x$$

Then,

$$A_{\mathrm{△}COB}=\frac{bh}{2}=\frac{\tan x}{2}$$

---

Therefore, we have

$$\begin{array}{rlr}{A}_{\mathrm{△}DOB}& \le A_{\text{sector }DOB}& \le A_{\mathrm{△}COB}\\ \frac{\sin x}{2}& \le \frac{x}{2}& \le \frac{\tan x}{2}\end{array}$$

Since $x\in (0,\frac{\pi }{2})$, we consider $x\to 0^{+}$.
Multiplying through by $\frac{2}{\sin x}>0$ yields

$$1\le \frac{x}{\sin x}\le \frac{1}{\cos x}$$

Taking reciprocals gives

$$\begin{array}{rlr}\cos x& \le \frac{\sin x}{x}& \le 1\\ ⟹\underset{x\to 0^{+}}{lim}\cos x& \le \underset{x\to 0^{+}}{lim}\frac{\sin x}{x}& \le \underset{x\to 0^{+}}{lim}1\\ 1& \le \underset{x\to 0^{+}}{lim}\frac{\sin x}{x}& \le 1\end{array}$$

$\therefore$ by the Squeeze Theorem,

$$\underset{x\to 0^{+}}{lim}\frac{\sin x}{x}=1$$

For $x\to 0^{-}$, since $\frac{\sin x}{x}$ is an even function as demonstrated below

$$\frac{\sin -x}{-x}=\frac{-\sin x}{-x}=\frac{\sin x}{x}$$

It follows that $\underset{x\to 0^{-}}{lim}\frac{\sin x}{x}=1$.
Hence,

$$\underset{x\to 0}{lim}\frac{\sin x}{x}=1$$

Derivative of a constant

[!abstract] $\frac{d}{dx}c=0$ for $c\in \mathbb{R}$

Let $f(x)=c$, then

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{c-c}{h}\\ & =\underset{h\to 0}{lim}\frac{0}{h}\\ & =\underset{h\to 0}{lim}0\\ & =0\end{array}$$

Derivative of function times a constant

[!abstract]+
Let $c\in \mathbb{R}$ and $f$ be a differentiable function, then

$$\frac{d}{dx}(cf(x))=c\frac{d}{dx}f(x)$$

Let $g(x)=cf(x)$, then

$$\begin{array}{rl}\frac{d}{dx}(cf(x))=g^{\prime }(x)& =\underset{h\to 0}{lim}\frac{g(x+h)-g(x)}{h}\\ & =\underset{h\to 0}{lim}cf(x+h)-f(x)\end{array}$$

Derivative of $\sin x$

Derivative of $\tan x$

$$\begin{array}{rl}\frac{d}{dx}\tan x& =\frac{d}{dx}\frac{\sin x}{\cos x}\\ & =\frac{\cos x(\sin x{)}^{\prime }-\sin x(\cos x{)}^{\prime }}{{\cos }^{2}x}\\ & =\frac{\cos x\cos x+\sin x\sin x}{{\cos }^{2}x}\\ & =\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}\\ & =\frac{1}{{\cos }^{2}x}\\ & ={\sec }^{2}x\end{array}$$

Derivative of $\cot x$

$$\begin{array}{rl}\frac{d}{dx}\cot x& =\frac{d}{dx}\frac{\cos x}{\sin x}\\ & =\frac{\sin x(\cos x{)}^{\prime }-\cos x(\sin x{)}^{\prime }}{{\sin }^{2}x}\\ & =\end{array}$$

Derivative of ${\log }_{b}x$

[!abstract] $\frac{\mathrm{d}}{\mathrm{d}x}{\log }_{b}x=\frac{1}{x\ln b}$

Note that

$$\underset{n\to \mathrm{\infty }}{lim}(1)$$

Let $f(x)={\log }_{b}x$, then

Derivative of $\arcsin x$

Derivative of $\arccos x$

Differentiability implies continuity

We have to show that $f(a)=\underset{x\to a}{lim}f(x)$. We know that $f$ is differentiable at $a$, so

$$f^{\prime }(a)=\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}$$

is certain to exist.
Then,

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{x\to a}{lim}\frac{f(x)-f(a)}{x-a}\\ f^{\prime }(x)& =\frac{\underset{x\to a}{lim}(f(x)-f(a))}{\underset{x\to a}{lim}(x-a)}\\ f^{\prime }(x)\cdot \underset{x\to a}{lim}(x-a)& =\underset{x\to a}{lim}(f(x)-f(a))\\ f^{\prime }(x)\cdot 0& =\underset{x\to a}{lim}f(x)-\underset{x\to a}{lim}f(a)\\ \underset{x\to a}{lim}f(a)& =\underset{x\to a}{lim}f(x)\\ f(a)& =\underset{x\to a}{lim}f(x)\end{array}$$

The contrapositive of the theorem says non-continuity $\implies$ non-differentiability.

The converse is not necessarily true, i.e. continuity doesn't imply differentiability.

Contributors

Ajitani Hifumi

Changelog

Last edited 3 minutes ago

View full history

779bb83-Vault update: 2024-10-16T19:38:06

8e28bb0-Vault update: 2024-10-10T01:00:24

b6f0df6-Vault update: 2024-10-09T08:30:52

fb67154-Vault update: 2024-10-07T19:40:37

6eaef0a-Vault update: 2024-10-01T18:02:12