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.
Definitions:
$y=e^x$ 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.