\(\underset { x\rightarrow \frac { 1 }{ 2 } }{ lim } \frac { { 4x }^{ 2 }-1 }{ 2x-1 } =\underset { x\rightarrow \frac { 1 }{ 2 } }{ lim } \frac { \left( { 2x }^{ 2 } \right) -1^{ 2 } }{ 2x-1 } =\underset { x\rightarrow \frac { \pi}{ 2 } }{ lim } \frac { (2x+1)(2x-1) }{ 2x-1 } =\underset { x\rightarrow \frac { \pi }{ 2 } }{ lim } (2x+1)\)
= \(2\left( \frac { 1 }{ 2 } \right) +1=1+1=2\)

Given f(x) = |x - 1|

\(f(x) = \begin{cases} x-1\quad if\quad x\ge 1 \\ 1-x\quad if\quad x<1 \end{cases}\)
\(L\left[ f\left( 1 \right) \right] =\underset { h\rightarrow 0 }{ lim } \frac { f(1-h)-f(1) }{ 1-h-1 } \left[ \therefore f(x)=1-xifx<1 \right] \)
\(=\underset { h\rightarrow 0 }{ lim } \frac { \left[ 1-(1-h)-[1-1] \right] }{ 1-h-1 } \)
\(=\underset { h\rightarrow 0 }{ lim } \frac { (1-1+h)-0 }{ -h } =\underset { h\rightarrow 0 }{ lim } \frac { h }{ -h } =-1\);...(1)
\(R\left[ f\left( 1 \right) \right] =\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ 1+h-1 } [\therefore f(x)-x-1ifx\ge 1]\)
\(=\underset { h\rightarrow 0 }{ lim } \frac { (1+h)-1-[1-1] }{ 1+h-1 } \)
= \(\underset { h\rightarrow 0 }{ lim } \frac { h-0 }{ h } =1\) ......(2)
From (1) and (2),
L[f '(1)]\(\neq \) \(R\left[ { f }^{ ' }\left( 1 \right) \right] \)
f(x) is not differentiable at x=1

Given f(x) = 5x - 3
\(L\left[ f\left( x \right) \right] _{ x=3 }\) =\(\underset { x\rightarrow 3 }{ lim } f(x)=\underset { h\rightarrow 0 }{ lim } f(3-h)\)
= \(\underset { h\rightarrow 0 }{ lim } 5(3-h)-3=\underset { h\rightarrow 0 }{ lim } (15-5h-3)\)
= \(\underset { h\rightarrow 0 }{ lim } \) (12-5h) = 12-0=12
\(R\left[ f\left( x \right) \right] _{ x=3 }=\underset { x\rightarrow 3^{ + } }{ lim } f(x)=\underset { h\rightarrow 0 }{ lim } f(3+h)\)
= \(\underset { h\rightarrow 0 }{ lim } 5(3+h)-3=\underset { h\rightarrow 0 }{ lim } 15+5h-3\)
= \(\underset { h\rightarrow 0 }{ lim } 12+5h=12-0=12\)
\(L\left[ f\left( x \right) \right] _{ x=3 }=R\left[ f\left( x \right) \right] _{ x=3 }\) 
ஃ f(x) is continous at x =3

Given \(f(x) =\begin{cases} \frac { 1-cos4x }{ 8{ x }^{ 2 } } \quad ifx\neq 0 \\ k\quad \quad \quad ifx=0 \end{cases}\)
\(\neq \underset { x\rightarrow 0 }{ lim } \quad f(x)=\underset { x\rightarrow 0 }{ lim } \frac { 1-cos4x }{ { 8x }^{ 2 } } =\underset { x\rightarrow 0 }{ lim } \frac { 2sin^{ 2 }2x }{ { 8x }^{ 2 } } \)   [ஃ 1-cos²z =sin²x]
= \(\underset { x\rightarrow 0 }{ lim } \frac { sin^{ 2 }2x }{ { 4x }^{ 2 } } =\underset { x\rightarrow 0 }{ lim } \left( \frac { sin2x }{ 2x } \right) ^{ 2 }\)
= (1)²    ....(1)
Given f(0) = k ...(2)
Since f(x) is continous at x = 0,
\(\underset { x\rightarrow 0 }{ lim } \) f(x) = f(0)
1 = k [using (1) and (2)
K = 1

Given f(x) = 5x - \(\left| x \right| \)
\(\therefore f(x)=\begin{cases} 5x-x\quad if\quad x\ge 0 \\ 5x-(-x)\quad if\quad x>0 \end{cases}=\begin{cases} 4x\quad ifx\ge 0 \\ 6x\quad if\quad x<0 \end{cases}\)
\(L\left[ f\left( x \right) \right] _{ x=0 }=\underset { x\rightarrow 0 }{ lim }f\left( x \right) =\underset { h\rightarrow 0- }{ lim } \quad f(o-h)\)
\(=\underset { h\rightarrow 0 }{ lim } f(-h)=\underset { h\rightarrow 0 }{ lim } 6(-h)\quad \quad \left[ \therefore f(x)=6x\quad ifx\ge 0 \right] \)
= 0
\(R\left[ f\left( x \right) \right] _{ x=0 }=\underset { x\rightarrow 0 }{ lim } lif\left( x \right) =\underset { h\rightarrow 0 }{ lim } \quad f(o+h)\)
\(\underset { h\rightarrow 0 }{ lim } f(h)=\underset { h\rightarrow 0 }{ lim } 4(h)\quad \left[ \therefore f(x)=4xi\quad fx\ge 0 \right] \)
=0
∴ \(L\left[ f\left( x \right) \right] _{ x=0 }=R\left[ f\left( x \right) \right] _{ x=0 }\)
∴ f(x) is continuous at x = 0