Discuss the continuity of the following function:
\[f\left( x \right)=sin\left( x \right).cos\left( x \right)\]

Hint: A function will be continuous if its left hand limit and right hand limit at a point will be equal to the value of the function at that point. As no specific point is mentioned in the question, we have to check continuity at all points, therefore we should take any arbitrary point c and check the continuity of the given function f(x)at c.

Complete step-by-step solution -
We should first understand the definitions of limit and continuity which are the following:
(a) Limit of a function f(x) at a point c is defined to be the value that the function approaches when we take x closer and closer to c.

Now, we can see that x can approach c from the right hand side in the number line, i.e. x is greater than c but gradually its value is reduced to become closer and closer to c. The limit obtained by approaching to c in this way is known as the right hand limit of f(x) at c and is denoted by $\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(x+h)$ where h is a small positive number.
Similarly, if the value of x is initially smaller than c and is gradually increased to approach c, the limit obtained in this way is known as the left hand limit of f(x) at c and is denoted by $\underset{x\to {{c}^{-}}}{\mathop{\lim }}\,f(x)=\underset{h\to 0}{\mathop{\lim }}\,f(c-h)$ where h is a small positive number.

(b) A function is said to be continuous if its left hand limit and right hand limit at a point will be equal to the value of the function at that point.
The given function f(x) can be written as $f(x)=\sin (x).\cos (x)=\dfrac{2\sin (x).\cos (x)}{2}=\dfrac{\sin (2x)}{2}$.
(as $\sin (a\pm b)=\sin (a)\cos (b)\pm \cos (a)\sin (b)....(1.1)$
$\sin (2x)=\sin (x+x)=\sin (x).\cos (x)+\cos (x).\sin (x)=2\sin (x).\cos (x)$)
At an arbitrary point c, the left hand limit of f(x) at c would be given by
$\begin{align}
  & \underset{h\to 0}{\mathop{\lim }}\,f(c-h)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (2(c-h))}{2}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (2c-2h)}{2} \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (2c)\cos (2h)-\cos (2c)\sin (2h)}{2}(\text{using equation (1}\text{.1)}) \\
 & =\dfrac{\sin (2c)}{2}\underset{h\to 0}{\mathop{\lim }}\,\cos (2h)-\dfrac{\cos (2c)}{2}\underset{h\to 0}{\mathop{\lim }}\,\sin (2h)=\dfrac{\sin (2c)}{2}\times 1-\dfrac{\cos (2c)}{2}\times 0\text{ }(\text{as sin(0)=0 and cos(0)=1}) \\
 & =\dfrac{\sin (2c)}{2}=f(c).........(1.2) \\
\end{align}$
And the right hand limit will be given by
$\begin{align}
  & \underset{h\to 0}{\mathop{\lim }}\,f(c+h)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (2(c+h))}{2}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (2c+2h)}{2} \\
 & =\underset{h\to 0}{\mathop{\lim }}\,\dfrac{\sin (2c)\cos (2h)+\cos (2c)\sin (2h)}{2}(\text{using equation (1}\text{.1)}) \\
 & =\dfrac{\sin (2c)}{2}\underset{h\to 0}{\mathop{\lim }}\,\cos (2h)+\dfrac{\cos (2c)}{2}\underset{h\to 0}{\mathop{\lim }}\,\sin (2h)=\dfrac{\sin (2c)}{2}\times 1+\dfrac{\cos (2c)}{2}\times 0\text{ }(\text{as sin(0)=0 and cos(0)=1}) \\
 & =\dfrac{\sin (2c)}{2}=f(c)........(1.3) \\
\end{align}$
Thus from equations (1.2) and (1.3), we see that,
Left hand limit of f(x) at c = Right hand limit of f(x) at c=f(c)
Thus, f is continuous at c. However, as c was any arbitrary point, f(x) is continuous at all points (this result is true for complex numbers as well because we did not use any property of real numbers in solving the question).

Note: The continuity could also be derived by directly taking the function (without converting it to sin(2x). However, due to multiplication, there would be a lot of terms in step 2 and 3 and thus would be difficult to simplify.