\(\lim_ { x\rightarrow 0 }{ lim } \frac { sin\quad x-2sin\quad 3x+sin\quad 5x }{ x } \)
\(=\lim_{ x\rightarrow 0 }{ lim } \frac { \left( sin5x\quad +\quad sin\quad x \right) -2sin\quad 3x }{ x } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { 2sin\left( \frac { 5x+x }{ 2 } \right) cos\left( \frac { 5x-x }{ 2 } \right) -2sin3x }{ x } \)
\( \left[ \because \ sin\ C+sin\ D=2sin\ \left( \frac { C+D }{ 2 } \right) COS\left( \frac { C-D }{ 2 } \right) \right] \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { 2sin3x\quad cos2x-2sin3x }{ x } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { 2sin3x(cos\quad 2x-1) }{ x } \)
\(=\lim_ { x\rightarrow 0 }{ lim } 2\left( \frac { sin\quad 3x }{ 3x } \right) \times 3\left[ \frac { -1(1-cos2x) }{ 1 } \right] \)
\(=-2\times 1\times 3(1-cos\quad 0) \quad \left[ \because \quad \underset { x\rightarrow 0 }{ lim } \frac { sin\quad \theta }{ \theta } \right] \)
\(=-6(1-1)=-6\times 0=0\)

\( \lim _{x \rightarrow \pi / 6} \frac{\cot ^{2} x-3}{\operatorname{cosec} x-2} \)
\(= \lim _{x \rightarrow \pi / 6} \frac{\operatorname{cosec}^{2} x-1-3}{\operatorname{cosec} x-2} \quad\left[\because \operatorname{cosec}^{2} x-\cot ^{2} x=1\right] \)
\(= \lim _{x \rightarrow \pi / 6} \frac{\operatorname{cosec}^{2} x-4}{\operatorname{cosec} x-2} \)
\(= \lim _{x \rightarrow \pi / 6} \frac{(\operatorname{cosec} x-2)(\operatorname{cosec} x+2)}{(\operatorname{cosec} x-2)} \quad[\text { by factorisation }] \)
\(=\lim _{x \rightarrow \pi / 6}(\operatorname{cosec} x+2)\)
\(= \operatorname{cosec} \frac{\pi}{6}+2=2+2=4\)

 \(f\prime (x)=\lim_ { h\longrightarrow 0 }{ lim } \left[ \frac { { \left( x+h \right) }^{ 2/3 }-{ x }^{ 2/3 } }{ h } \right] \)
\(=\lim_ { (x+h)\longrightarrow x }{ lim } \left[ \frac { { (x+h) }^{ 2/3 }-{ x }^{ 2/3 } }{ (x+h)-x } \right] \)
\(=\frac { 2 }{ 3 } { x }^{ -1/3 }\)

\(\lim_{ h\rightarrow 0 } \frac { (a+h)^{ 2 }sin(a+h)-a^{ 2 }sin\quad a }{ h } \)
\(=\lim_ { lim }{ h\rightarrow 0 } \frac { (a^{ 2 }+h^{ 2 }+2ah)[sin \ a\ cos \ h+cos\ a\ sin \ h)-a^{ 2 }sin\quad a }{ h } \)
\(\left[ \because \ sin(C+D)=sinCcosD+CosCsinD \right] \)
\(=\lim_{ h\rightarrow 0 } \left[ \frac { a^{ 2 }sin \ a(cos \ h-1) }{ h } +\frac { a^{ 2 }cos \ a \ sin \ h) }{ h } +(h+2a)(sin \ a \ cos \ h+cos\ a \ sin\ h) \right] \)
\(=\lim_{ h\rightarrow 0 } \left[ \frac { a^{ 2 }sina(-2sin^{ 2 }\frac { h }{ 2 } ) }{ \frac { h^{ 2 } }{ 2 } } .\frac { h }{ 2 } \right] +\overset { lim }{ h\rightarrow 0 } \frac { a^{ 2 }cosasinh }{ h } +\lim_{ h\rightarrow 0 } (h+2a)sin(a+h)\)
\(\left[ \because cosm \ h-1=-2sin^{ 2 } h/2\quad and\quad sinacos\quad h+cosasin\quad h=sin(a+h) \right] \)
\(=a^{ 2 }sin\quad a\times 0+a^{ 2 }\quad cosa(1)+2asina \quad \left[ \because \lim_{ x\rightarrow 0 } \frac { sin\quad x }{ x } =1 \right]\)
\( =a^{ 2 }cos \ a+2asin \ a\)

Given limit = \(\lim_ { x\rightarrow 0 }{ lim } \frac { \sqrt { a+x } -\sqrt { a } }{ x\sqrt { { a }^{ 2 }+ax } } \times \frac { \sqrt { a+x } +\sqrt { a } }{ \sqrt { a+x } +\sqrt { a } } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { x }{ x(\sqrt { { a }^{ 2 }+ax } )(\sqrt { a+x } +\sqrt { a } ) }\)
\(\frac { 1 }{ 2a\sqrt { a } } \)

\(\lim_ { x\rightarrow 2 }{ lim } \frac { { 3x }^{ 2 }-x-10 }{ { x }^{ 2 }-4 } =\lim_ { x\rightarrow 2 }{ lim } \frac { (x-2)(3x+5) }{ (x-2)(x+2) } \)
\(\frac { 11 }{ 4 } \)

Given limit = \(\lim_ { x\rightarrow 0 }{ lim } \left[ \frac { tanx }{ x } -\frac { sinx }{ x } \right] \)
=0

\(\lim _{x \rightarrow 1} \frac{x^{3}-1}{x-1}=\lim _{x \rightarrow 1} \frac{(x-1)\left(x^{2}+x+1\right)}{(x-1)}\)
3

\(\lim _{x \rightarrow 0} \frac{\cos a x-\cos b x}{\cos \alpha x-1}\)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { -2sin\left[ \left( \frac { a+b }{ 2 } \right) x \right] sin\left( \frac { a-b }{ 2 } \right) x }{ -2{ sin }^{ 2 }\frac { cx }{ 2 } } \)
\(\left[ \because \ cos \ C-cos \ D=-2sin\left( \frac { C+D }{ 2 } \right) sin\left( \frac { C-D }{ 2 } \right) and \ cos\ 2\theta =1-2{ sin }^{ 2 }\theta \right] \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { sin\left( \frac { a+b }{ 2 } \right) x.sin\left( \frac { a-b }{ 2 } \right) x }{ x^{ 2 } } .\frac { { x }^{ 2 } }{ { sin }^{ 2 }\frac { cx }{ 2 } } \)
\(=\lim_ { x\rightarrow 0 }{ lim } \frac { sin\left( \frac { a+b }{ 2 } \right) x }{ \left( \frac { a+b }{ 2 } \right) x.\left( \frac { 2 }{ a+b } \right) } .\frac { sin\left( \frac { a-b }{ 2 } \right) x }{ \left( \frac { a-b }{ 2 } \right) x.\left( \frac { 2 }{ a-b } \right) } .\frac { { \left( \frac { cx }{ 2 } \right) }^{ 2 }\times \frac { 4 }{ { c }^{ 2 } } }{ { sin }^{ 2 }\frac { cx }{ 2 } } \)
\( =\left( \frac { a+b }{ 2 } \times \frac { a-b }{ 2 } \times \frac { 4 }{ { c }^{ 2 } } \right) =\frac { { a }^{ 2 }-{ b }^{ 2 } }{ { c }^{ 2 } } \left[ \because \quad \lim_ { x\rightarrow 0 }{ lim } \frac { sin\quad x }{ x } =1 \right] \)

Given \(\lim _{ x\rightarrow 4 }{ \frac { |x-4| }{ x-4 } } \)
\(LHL=\lim _{ { X\rightarrow 4 }^{ 1 } }{ \frac { -(x-4) }{ x-4 } } =1 \quad \left[ \because |x-4|=-(x-4),x<4 \right] \)
\(RHL=\lim _{ { X\rightarrow 4 }^{ + } }{ \frac { (x-4) }{ x-4 } } =1\quad \left[ \because |x-4|=(x-4),x<4 \right] \)

\(LHL=\lim _{ X\rightarrow { -1 }^{ - } }{ (x+2)= } \lim _{ h\rightarrow 0 }{ (-1-h+2)=1 } \)
\(RHL=\lim _{ x\rightarrow { 1 }^{ + } }{ f(x) } =\lim _{ x\rightarrow { -1 }^{ + } }{ { cx }^{ 2 } } =\lim _{ h\rightarrow 0 }{ c{ (-1+h })^{ 2 } } \)
Ans. c = 1

   \(=\frac { \lim _{ x\rightarrow 0 }{ \frac { \left( 1+5 \right) ^{ 2 }-1 }{ x } } }{ \lim _{ x\rightarrow 0 }{ \frac { 3x+{ 5 }x^{ 2 } }{ x } } } =\frac { \lim _{ (1+x)\rightarrow 1 }{ \frac { (1+x)^{ 5 }-1 }{ (1+x)-1 } } }{ \lim _{ x\rightarrow 0 }{ (3x+5x) } } \)
   Ans \(\frac { 5 }{ 3 } \)

\(Given,\ y=\frac { sinx+cosx }{ sinx-cosx } \)
\(\therefore \frac { \left[ (sinx-cosx)(cosx-sinx)-(sin+cosx)(cosx+sinx) \right] }{ \left( sinx-cosx \right) ^{ 2 } } \)
\( =\frac { -(sinx-cosx)^{ 2 }-(sinx+cosx)^{ 2 } }{ (sinx-cosx)^{ 2 } } =-2\)

\(LHL=\lim _{ x\rightarrow { 1 }^{ - } }{ a+bx)= } \lim _{ h\rightarrow { 0 } }{ [a+b(1-h)]=a+b } \)
\(RHL=\lim _{ x\rightarrow { 1 }^{ + } }{ (b-ax) } =\lim _{ h\rightarrow { 0 } }{ [b-a(1+h)]=b-a } \)
\(\because LHL=RHL=f\left( 1 \right) \Rightarrow A+B=B-A=4\)
Ans. A = 0, B = 4

\(LHL=\lim _{ x\rightarrow { 1 }^{ - } }{ (5x-4) } =\lim _{ h\rightarrow 0 }{ [5(1-h)-4]=1 } \) and 
\(RHL=\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } =\lim _{ x\rightarrow { 1 }^{ + } }{ { (4x }^{ 2 }-3x) } \)
\(=\lim _{ h\rightarrow { 0 } }{ [4(1+{ h })^{ 2 } } -3(1+h)]=1\)
\(\lim _{ x\rightarrow { 1 }^{ + } }{ f\left( x \right) } \) exists and it is equal to 1