Continuous function vs smooth function I don't quite get the difference between the continuous function and the smooth function. Moreover, are they equivalent, or if a function is smooth, it is continuous, or vice versa?
 A: 
Moreover, are they equivalent, or if a function is smooth, it is continuous, or vice versa?

If a function is smooth (or even just once differentiable) at a point, then it must be continuous at that point as well.  If the simple limit required for continuity failed to exist, then the more complicated limit required for differentiability would also fail to exist.
However, as the following example demonstrates, the converse does not hold, so they are not equivalent.
Consider the absolute value function, $f(x) = |x|$.
A function is continuous at a point $x_0$ if $$
\lim_{x\rightarrow x_0} f(x) = f(x_0)
$$
So we need to evaluate the limit for $|x|$.  We have:$$
|x| = 
\begin{cases}
\hphantom{-}x, & x > 0 \\
-x, & x < 0 \\
\hphantom{-}0, & x = 0
\end{cases}
$$
$x$ and $-x$ are both "obviously" continuous, and when you plug them into the limit expression, they clearly converge:$$
\lim_{x\rightarrow x_0} x = x_0, \: x_0 > 0 \\
\lim_{x\rightarrow x_0} -x = -x_0, \: x_0 < 0
$$
That leaves us with $x_0 = 0$.  The limit exists if and only if both one-sided limits exist and are equal to each other.  So consider:$$
\lim_{x\rightarrow 0^-} |x| = \lim_{x\rightarrow 0^-} -x = 0 \\
\lim_{x\rightarrow 0^+} |x| = \lim_{x\rightarrow 0^+} x = 0
$$
The one-sided limits are equal to each other, and to the actual absolute value function, at zero.  So the function is continuous at zero (and therefore everywhere).
A function is smooth at a point $x$ if its $n$th derivative exists at that point, for all positive integers $n$.  If this doesn't work even for $n=1$, then it can't possibly work for all $n$, so take the first derivative of $|x|$.  For $x \neq 0$ it's straightforward enough:$$
f'(x) =\lim_{h \rightarrow 0} \frac{|x + h| - |x|}{h} \\
f'(x) =\begin{cases}
\lim_{h \rightarrow 0} \frac{x + h - x}{h}, & x > 0\\
\lim_{h \rightarrow 0} \frac{- x - h + x}{h}, & x < 0\\
\end{cases} \\
f'(x) = \begin{cases}
\hphantom{-}1, & x > 0 \\
-1, & x < 0
\end{cases}
$$
(Note that as $h\rightarrow 0$, it will eventually become small enough that $x$ and $x + h$ necessarily have the same sign, regardless of the direction of the limit.)
But it all blows up if $x = 0$:$$
f'(0) =\lim_{h \rightarrow 0} \frac{|0 + h| - |0|}{h} \\
f'(0) =\lim_{h \rightarrow 0} \frac{|h|}{h} \\
f'(0) = \begin{cases}
\lim_{h \rightarrow 0^-} \frac{-h}{h} \\
\lim_{h \rightarrow 0^+} \frac{h}{h}
\end{cases} \\
f'(0) = \begin{cases}
-1 \\
\hphantom{-}1
\end{cases}
$$
Since we can't have both $f'(0) = -1$ and $f'(0) = 1$, this function is not differentiable (and therefore not smooth) at $x = 0$.
