I want to prove that $\exp x$ and $\sin x$ are continuous. This means I want to show that

$$\lim\limits_{x\to a}e^x=e^a$$

$$\lim\limits_{x\to a}\sin x=\sin a$$

for any fixed $a \in \Bbb R$. Then I need to show that for any $\epsilon >0$ there exists a $\delta >0$ such that whenever $|x-a|<\delta$, then $|\sin x -\sin a|<\epsilon$ and similarily $|e^x-e^a|<\epsilon$. For the first case I have that

$$\left| {\sin x - \sin a} \right| = 2\left| {\sin \frac{{x - a}}{2}} \right|\left| {\cos \frac{{x + a}}{2}} \right|$$

for the second case I use

$$\left| {{e^x} - {e^a}} \right| = \left| {{e^a}} \right|\left| {{e^{x - a}} - 1} \right|$$

ADD Just for the sake of a simple solution:

Let $\epsilon>0$ and $a\in \bf R$ be given. Choose $\delta = \epsilon$. Then we have that

$$\eqalign{ \left| {\sin x - \sin a} \right| = 2\left| {\sin \left( {\frac{{x - a}}{2}} \right)\cos \left( {\frac{{x + a}}{2}} \right)} \right| &\cr = 2\sin \left| {\frac{{x - a}}{2}} \right|\cos \left| {\frac{{x + a}}{2}} \right| &\cr \leqslant 2\left| {\frac{{x - a}}{2}} \right| = \left| {x - a} \right| < \delta = \epsilon &\cr} $$

For the exponential, we have that $\log:(0,+\infty)\to \Bbb R$ is continuous, one-one, onto, and differentiable on $(0,+\infty)$. It follows that $\exp:\mathbb R\to (0,\infty)$ defined by $\exp x =y \iff \log y = x$ is also continuous and differentiable.


$y=e^x$ is the inverse of $y=\log x $ defined as

$$\log x = \lim_{k \to 0} \frac{x^k-1}{k}$$

Here I gave a list of the elementary functions of $\log$.

I would probably choose to define $\sin$ like this

Checking Apostol's Calculus I found he proves the continuity of $\sin x$ by first proving that if $f$ is (Riemann) integrable in $[a,x]$ for all $x\in [a,b]$ then

$$F(x) = \int_a^x f(t) dt$$

is continuous for all $x\in [a,b]$. Under this light it is evident

$$\sin x = \int_0^x \cos t dt $$ $$\cos x = 1-\int_0^x \sin t dt $$ $$e^x = \int_0^x e^t dt+1$$

are all continuous.

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    $\begingroup$ This will probably depend a lot of what your definitions of $\exp$ and $\sin$ are; there are several common ones that demand quite different proofs for fundamental properties such as these. Could you add a summary of your definitions and which properties you have already established? $\endgroup$ – hmakholm left over Monica Apr 25 '12 at 22:08
  • $\begingroup$ @Henning Done. space filling text $\endgroup$ – Pedro Tamaroff Apr 25 '12 at 22:24
  • $\begingroup$ I think in both cases you are in position to use the inverse function theorem for strictly monotone functions (Theorem 3.2.5 in those notes). I linked you to it right before you deleted your last question, so I'm not sure if you saw the comment. $\endgroup$ – t.b. Apr 25 '12 at 22:26
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    $\begingroup$ @Peter: Protip: you can use ${}{}{}{}{}{}$ as invisible filler in short comments. $\endgroup$ – hmakholm left over Monica Apr 25 '12 at 22:26
  • $\begingroup$ @copper.hat, I deleted the comment because I realized it was circular. $\endgroup$ – Nicholas Stull Apr 26 '12 at 1:14

There's a sense of deja vu here. I'm not exactly what you consider axiomatic.

Here's a different approach for $\exp$.

The function $x \mapsto \exp(x)$ is, by your definition, the inverse of $x \mapsto \log(x)$. The function $\log$ is continuous, strictly increasing, and the range is $\mathbb{R}$. Hence it follows that $\exp$ is strictly increasing and defined on all of $\mathbb{R}$. Let $\epsilon>0$, and let $\delta = \min\{-\log(1-\epsilon), \log(1+\epsilon)\}$, and suppose $|x| < \delta$. Then since $$\log(1-\epsilon) < x < \log(1+\epsilon)$$ we have $$ 1-\epsilon = \exp(\log(1-\epsilon)) < \exp(x) < \exp(\log(1+\epsilon)) = 1+\epsilon$$ Hence $|\exp(x)-1| < \epsilon$, and so $\exp$ is continuous at $0$.

Original 'non-answer':

For $\sin$, you know that $\sin^2(x)+\cos^2(x) = 1$, hence $\cos(x)$ is bounded by $1$. Since $\cos$ is the derivative of $\sin$, the mean value theorem shows that $|\sin(x)-\sin(y)| \leq |x-y|$ for any $x,y$.

Replacement answer:

In context of the definition of $\sin$ from the question: The function $x \mapsto A(x)$ is strictly increasing and continuous on $[-1,1]$. Using a similar argument to that used for $\exp$ above, the function $\cos$ can be seen to be continuous. Hence, since $\sin(x) = \sqrt{1-\cos^2(x)}$, it follows that $\sin$ is continuous.

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  • $\begingroup$ I'm thinking there is perhaps an interesting way to show this by noting that $|\sin(x)| \leq |x|$ $\endgroup$ – Nicholas Stull Apr 26 '12 at 1:15
  • $\begingroup$ @NicholasStull: How do you show this in an elementary fashion without the MVT? $\endgroup$ – copper.hat Apr 26 '12 at 1:17
  • $\begingroup$ Another thing to consider is this: If we prove that the exponential function is continuous (in $\mathbb{C}$, unfortunately), use Euler's formula: $$\sin(x) = \frac{1}{2i}(e^{ix} - e^{-ix})$$ $\endgroup$ – Nicholas Stull Apr 26 '12 at 1:18
  • $\begingroup$ @copper.hat, I'm trying to think of how to prove this without the MVT, and outside of using the scapegoat "look at the graph" method, I'm not sure yet. $\endgroup$ – Nicholas Stull Apr 26 '12 at 1:19
  • $\begingroup$ @Nicholas, perhaps this is what you're looking for? $\endgroup$ – Antonio Vargas Apr 26 '12 at 1:22

For the case of continuity of SINX, after what you made, I would use that absolute value of SIN(u(x))

is lower or equal than absolute value of U(X), and absolute value of cos(U(X)) is lowre or equal to 1.

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  • $\begingroup$ U(X) is any function where you apply sin, composition of sinx with any function.I don't have LATEX, sorry.In your case U(x)=(x-a)/2 for sin and (x+a)/s for cos $\endgroup$ – alpha.Debi Apr 28 '12 at 20:05

Proof that both $sin(x)$ and $cos(x)$ are uniformly continuous on $[0, 2 \pi]$.

Let $I = [0, 2 \pi]$ be subdivided into $2^n$ subintervals, as follows

$x_{k} = k \frac{2 \pi}{2^n}\;$ for $0 \le k \le 2^n$.

Define $z_0 = 1$, $z_1 = -1$, $z_2 = i$, and

$z_{m+1} = \sqrt {z_m}$, with $Re(z_{m+1}) > 0$ for $m \ge 2$

For each $n$ define the discrete function $W_n$ as

$k \frac{2 \pi}{2^n}\; \mapsto z_n^k$, $\,0 \le k \le 2^n$.

sending a finite number of points into the complex circle.

Define the $cos$ and $sin$ functions as follows:

$z_n^k = cos(k \frac{2 \pi}{2^n}) + sin(k \frac{2 \pi}{2^n})i$

The graphs of the functions $W_n$ satisfy $W_n \subset W_{n+1}$, so we can take the union of these sets and get a new function $W$ defined on a dense subset $D$ of $I$.

Proposition 1: The function $W$ is uniformly continuous on $D$, taking countably infinite values on the unit circle. Both $sin(x)$ and $cos(x)$ are also uniformly continuous.

Proof: Compare the lengths of chords and arcs on the unit circle.

Proposition 2: The function $W$ is uniformly continuous on $I$, with range the unit circle. Both $sin(x)$ and $cos(x)$ are also uniformly continuous.

Proof: Extend by continuity.

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