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Question: Show that the function f(x) defined by \[f(x) = \left\\{ {\begin{array}{*{20}{c}} {\dfrac{{\si...

Show that the function f(x) defined by

{\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 $x=0$.
Explanation

Solution

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

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

Given f(x)f(x) is 22 at x=0x = 0i.e., f(0)=2f(0) = 2. Hence f(x)f(x) is defined at x=0x = 0
Consider LHS=f(0−)LHS = f(0 - )
By the definition of left-hand limit, we get
f(0−)=lim⁡h→0f(0−h)f(0 - ) = \mathop {\lim }\limits_{h \to 0} f(0 - h)-------(1)
From the given function we have f(x)=4(1−1−x)xf(x) = \dfrac{{4\left( {1 - \sqrt {1 - x} } \right)}}{x} when x<0x < 0, then the equation (1) becomes
f(0−)=lim⁡h→04(1−1−(−h))(−h)f(0 - ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{4\left( {1 - \sqrt {1 - ( - h)} } \right)}}{{( - h)}}
\Rightarrow $$$$f(0 - ) = - 4 \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {1 - \sqrt {1 + h} } \right)}}{h} (00  form)\left( {\dfrac{0}{0}\;form} \right)
Hence by using L-hospital rule, we get
f(0−)=−4×lim⁡h→0(0−121+h)1f(0 - ) = - 4 \times \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {0 - \dfrac{1}{{2\sqrt {1 + h} }}} \right)}}{1}
\Rightarrow $$$$f(0 - ) = - 4 \times \left( { - \dfrac{1}{2}} \right) = 2
\Rightarrow $$$$LHS = 2-----(2)

RHS=f(0+)RHS = f(0 + )
By the definition of Right-hand limit, we get
RHS=f(0+)=lim⁡h→0f(0+h)RHS = f(0 + ) = \mathop {\lim }\limits_{h \to 0} f(0 + h)----(3)
From the given function we have f(x)=sin⁡xx+cos⁡xf(x) = \dfrac{{\sin x}}{x} + \cos x when x>0x > 0, then the equation (3) becomes
f(0+)=lim⁡h→0(sin⁡(0+h)0+h+cos⁡(0+h)))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+)=lim⁡h→0(sin⁡(h)h)+lim⁡h→0(cos⁡(h))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 lim⁡x→0sin⁡xx=1\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1, then the equation (5) becomes
f(0+)=1+cos⁡(0)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)=2LHS = RHS = f(0) = 2
So, f(x)f(x) is continuous at x=0x = 0. Hence proved.

Note:
Note that the function f(x)f(x) is said to be continuous on any set GG if it is continuous at each point of that set GG. 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.