Show that the function f(x) defined by
\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {\dfrac{{\sin x}}{x} + \cos x} \\
  2 \\
  {\dfrac{{4\left( {1 - \sqrt {1 - x} } \right)}}{x}}
\end{array}\;} \right.\begin{array}{*{20}{c}}
  {if\;x > 0} \\
  {if\;x = 0} \\
  {if\;x < 0}
\end{array}\] is continuous at $x=0$.

Hint: First we have to check the given function \[f(x)\] is defined at the given point \[x = a\]. If it is defined then find the both left-hand limit \[\left( {LHS} \right)\] and right-hand limit \[\left( {RHS} \right)\][If it exists].
If \[LHS = RHS = f(a)\] then the given function \[f(x)\] is continuous at \[x = a\].

Complete step by step solution:
A function \[f(x)\] is said to be continuous at \[x = a\]if it is satisfying the following
(i) \[f(x)\] exists at \[x = a\] i.e., \[f(a)\] exists.
(ii) Both \[f(a - )\]\[\left( {LHS} \right)\] and \[f(a + )\]\[\left( {RHS} \right)\]exists
Where \[f(a - ) = \mathop {\lim }\limits_{h \to 0} f(0 - h)\], by taking \[f(x)\]when\[x < 0\].
Where \[f(a + ) = \mathop {\lim }\limits_{h \to 0} f(0 + h)\], by taking \[f(x)\]when\[x > 0\].
(iii) \[LHS = RHS = f(a)\].

Given \[f(x)\] is \[2\] at \[x = 0\]i.e., \[f(0) = 2\]. Hence \[f(x)\] is defined at \[x = 0\]
Consider \[LHS = f(0 - )\]
By the definition of left-hand limit, we get
\[f(0 - ) = \mathop {\lim }\limits_{h \to 0} f(0 - h)\]-------(1)
From the given function we have \[f(x) = \dfrac{{4\left( {1 - \sqrt {1 - x} } \right)}}{x}\] when \[x < 0\], then the equation (1) becomes
\[f(0 - ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{4\left( {1 - \sqrt {1 - ( - h)} } \right)}}{{( - h)}}\]
\[ \Rightarrow \]\[f(0 - ) = - 4 \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {1 - \sqrt {1 + h} } \right)}}{h}\] \[\left( {\dfrac{0}{0}\;form} \right)\]
Hence by using L-hospital rule, we get
\[f(0 - ) = - 4 \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {0 - \dfrac{1}{{2\sqrt {1 + h} }}} \right)}}{1}\]
\[ \Rightarrow \]\[f(0 - ) = - 4 \times \left( { - \dfrac{1}{2}} \right) = 2\]
\[ \Rightarrow \]\[LHS = 2\]-----(2)

\[RHS = f(0 + )\]
By the definition of Right-hand limit, we get
\[RHS = f(0 + ) = \mathop {\lim }\limits_{h \to 0} f(0 + h)\]----(3)
From the given function we have \[f(x) = \dfrac{{\sin x}}{x} + \cos x\] when \[x > 0\], then the equation (3) becomes
\[f(0 + ) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{\sin (0 + h)}}{{0 + h}} + \cos \left( {0 + h)} \right)} \right)\]---(4)
We know that limit of sum of two functions is equal to the sum of the limits of the functions, then the equation (4) becomes
\[f(0 + ) = \mathop {\lim }\limits_{h \to 0} \left( {\dfrac{{\sin (h)}}{h}} \right) + \mathop {\lim }\limits_{h \to 0} \left( {\cos (h)} \right)\]---(5)
Since we know that \[\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1\], then the equation (5) becomes
\[f(0 + ) = 1 + \cos (0)\]
\[ \Rightarrow \]\[f(0 + ) = 2\]
\[ \Rightarrow \]\[RHS = 2\]---(6)
Hence from the equations (1) and (6), we get
\[LHS = RHS = f(0) = 2\]
So, \[f(x)\] is continuous at \[x = 0\]. Hence proved.

Note:
Note that the function \[f(x)\] is said to be continuous on any set \[G\] if it is continuous at each point of that set \[G\]. Also note that graphically, a function is continuous if it is drawn without lifting a pen. Every differentiable function is continuous but the converse is not true.