\(\underset { x\rightarrow 0 }{ lim } \frac { sin3x }{ sin5x } =\underset { x\rightarrow 0 }{ lim } \left\{ \frac { 3x\times \frac { sin3x }{ 3x } }{ 5x\times \frac { sin5x }{ 5x } } \right\} \)
\(=\frac { 3 }{ 5 } \frac { \underset { x\rightarrow 0 }{ lim } \frac { sin3x }{ 3x } }{ \underset { x\rightarrow 0 }{ lim } \frac { sin5x }{ 5x } } \)
\(=\frac { 3 }{ 5 } \left( \frac { 1 }{ 1 } \right) =\frac { 3 }{ 5 } \)

\(\underset { x\rightarrow a }{ lim } \frac { { x }^{ \frac { 3 }{ 5 } }-a^{ \frac { 3 }{ 5 } } }{ { x }^{ \frac { 1 }{ 5 } }-{ a }^{ \frac { 1 }{ 5 } } } =\underset { x\rightarrow a }{ lim } \frac { \frac { { x }^{ \frac { 3 }{ 5 } }-a^{ \frac { 3 }{ 5 } } }{ x-a } }{ \frac { { x }^{ \frac { 1 }{ 5 } }-{ a }^{ \frac { 1 }{ 5 } } }{ x-a } } \)
\(=\frac { \frac { 3 }{ 5 } \left( a \right) ^{ -\frac { 2 }{ 5 } } }{ \frac { 1 }{ 5 } \left( a \right) ^{ -\frac { 4 }{ 5 } } } \)
\(=3\left( a \right) ^{ -\frac { 2 }{ 5 } +\frac { 4 }{ 5 } }=3{ a }^{ \frac { 2 }{ 5 } }\)

\(f\left( x \right) =\frac { x-1 }{ x+1 } \)(கொடுக்கப்பட்டது)
\(\therefore \)     \(f\left[ f\left( x \right) \right] =\frac { f\left( x \right) -1 }{ f\left( x \right) +1 } =\frac { \frac { x-1 }{ x+1 } -1 }{ \frac { x-1 }{ x+1 } +1 } \)
\(=\frac { x-1-x-1 }{ x-1+x+1 } =-\frac { 2 }{ 2x } =-\frac { 1 }{ x } \)

\(f\left( x \right) =x+\frac { 1 }{ x } \),\(f\left( x^{ 3 } \right) ={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } ,f\left( \frac { 1 }{ x } \right) =f\left( x \right) \)
இப்போது, LHS = [f(x)]³
\(=\left[ x+\frac { 1 }{ x } \right] ^{ 3 }\)
\(={ x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } +3\left( x+\frac { 1 }{ x } \right) \)
\(=f\left( x^{ 3 } \right) +3f\left( x \right) =f\left( x^{ 3 } \right) +3f\left( \frac { 1 }{ x } \right) \)
= RHS

\(\underset { x\rightarrow \frac { \pi }{ 4 } }{ lim } \frac { 5sin2x-2cos2x }{ 3cos2x+2sin2x } =\frac { \underset { x\rightarrow \frac { \pi }{ 4 } }{ lim } \left( 5sin2x-2cos2x \right) }{ \underset { x\rightarrow \frac { \pi }{ 4 } }{ lim } \left( 3cos2x+2sin2x \right) } \)
\(=\frac { 5sin\frac { \pi }{ 2 } -2cos\frac { \pi }{ 2 } }{ 3cos\frac { \pi }{ 2 } +2sin\frac { \pi }{ 2 } } \)
\(=\frac { 5\left( 1 \right) -2\left( 0 \right) }{ 3\left( 0 \right) +2\left( 1 \right) } =\frac { 5 }{ 2 } \)

\(\underset { x\rightarrow 2 }{ lim } \frac { { x }^{ 2 }-4x+6 }{ x+2 } =\frac { \underset { x\rightarrow 2 }{ lim } \left( { x }^{ 2 }-4x+6 \right) }{ \underset { x\rightarrow 2 }{ lim } \left( x+2 \right) } \)
\(=\frac { \left( 2 \right) ^{ 2 }-4\left( 2 \right) +6 }{ 2+2 } =\frac { 1 }{ 2 } \)

\(\underset { x\rightarrow 1 }{ lim } \left( 3{ x }^{ 2 }+4x-5 \right) \)
= 3(1)² + 4(1) – 5 = 2

(f – g) (x) = f(x) – g(x) = x – |x|
\(=\begin{cases}x-x, x\ge0 \\x-(-x),x<0 \end{cases}=\begin{cases}0,x\ge0 \\ 2x,x<0 \end{cases}\)
 

(f + g) (x) = f(x) + g(x) = x + |x|
\(=\begin{cases}x+x,x\ge0\ \\x-x,x<0 \end{cases}=\begin{cases}2x,x\ge0 \\ 0,x<0 \end{cases}\)

\(f\left( x \right) =g\left( x \right) ,\forall x\in R\)(கொடுக்கப்பட்டது)
\(\therefore\)            f(–5) = g(–5)
–5 –5 = \(\lambda \)
\(\therefore\)         \(\lambda =-10\)