$$\eqalign{ & {A_i} = \frac{{x - {a_i}}}{{\left| {x - {a_i}} \right|}},\,i = 1,\,2,\,3,.....,\,n \cr & {a_1} < {a_2} < {a_3} < ..... < {a_n} \cr} $$
If $$x$$ is in the left neighborhood of $${a_1} < {a_2} < ..... < {a_{m - 1}} < x < {a_m} < {a_{m + 1}} < ..... < {a_n}$$
$$\eqalign{ & {A_i} = \frac{{x - {a_i}}}{{x - {a_i}}} = 1,\,i = 1,\,2,.....,\,m - 1\,; \cr & {A_i} = \frac{{x - {a_i}}}{{\left( {{a_i} - x} \right)}} = - 1,\,i = m,\,m - 1,.....,\,n \cr & \therefore \,{A_1}{A_2}.....{A_n} = {\left( { - 1} \right)^{n - m + 1}}......({\text{i}}) \cr} $$
If $$x$$ is in the left neighborhood of $${a_m},\,{a_1} < {a_2} < ..... < {a_{m - 1}} < {a_m} < x < {a_{m + 1}} < ..... < {a_n},$$
$$\eqalign{ & {A_i} = \frac{{x - {a_i}}}{{x - {a_i}}} = 1,\,i = 1,\,2,.....,\,n \cr & \therefore \,{A_1}{A_2}.....{A_n} = {\left( { - 1} \right)^{n - m}}......({\text{ii}}) \cr & \therefore \,\mathop {\lim }\limits_{x \to a_m^ - } \left( {{A_1}{A_2}.....{A_n}} \right) = {\left( { - 1} \right)^{n - m + 1}} \cr & {\text{and }}\mathop {\lim }\limits_{x \to a_m^ + } \left( {{A_1}{A_2}.....{A_n}} \right) = {\left( { - 1} \right)^{n - m}} \cr & \therefore \,L.H.L. \ne R.H.L. \cr & {\text{Hence, }}\mathop {\lim }\limits_{x \to {a_m}} \left( {{A_1}{A_2}.....{A_n}} \right){\text{ does not exist}}{\text{.}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {1 - \cos \,x} }} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {1 - \left( {1 - 2\,{{\sin }^2}\frac{x}{2}} \right)} }} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{x}{{\sqrt {2\,{{\sin }^2}\frac{x}{2}} }} \cr & = \frac{1}{2}\mathop {\lim }\limits_{x \to 0} \frac{x}{{\left| {\sin \frac{x}{2}} \right|}} \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( {0 - 0} \right) = \mathop {\lim }\limits_{h \to 0} \frac{x}{{\left| {\sin \frac{x}{2}} \right|}} \cr & = - \frac{1}{{\sqrt 2 }}\mathop {\lim }\limits_{x \to 0} \frac{{2\left( {\frac{h}{2}} \right)}}{{\sin \frac{h}{2}}} \cr & = \frac{1}{{\sqrt 2 }} \times 2 \times 1\,\,\,\left( {\because \,\mathop {\lim }\limits_{\theta \to 0} \frac{\theta }{{\sin \,\theta }} = 1} \right) \cr & = \sqrt 2 \cr & {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = f\left( {0 + 0} \right) = \mathop {\lim }\limits_{h \to 0} f\left( {0 + h} \right) \cr & = \frac{1}{{\sqrt 2 }}\mathop {\lim }\limits_{h \to 0} \frac{{2\left( {\frac{h}{2}} \right)}}{{\sin \frac{h}{2}}} \cr & = \frac{1}{{\sqrt 2 }} \times 2 \times 1 \cr & = \sqrt 2 \cr & {\text{L}}{\text{.H}}{\text{.L}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} = \sqrt 2 \cr & {\text{Therefore limit does not exist}}{\text{.}} \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{x \to 0} \frac{{2f'\left( x \right) - 6f'\left( {2x} \right) + 4f'\left( {4x} \right)}}{{2x}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{2f''\left( x \right) - 12f''\left( {2x} \right) + 16f''\left( {4x} \right)}}{2} \cr & = \frac{{6f''\left( 0 \right)}}{2} \cr & = 6 \cr} $$

$$\eqalign{ & {\text{Limit}} = \mathop {\lim }\limits_{n \to \infty } = \frac{{{{\sin }^2}\left( {n!} \right)}}{{{n^{1 - p}}\left( {1 + \frac{1}{n}} \right)}} \cr & = \frac{{{\text{some number between 0 and 1}}}}{\infty } \cr & = 0 \cr} $$

$$\eqalign{ & {\text{Given equation,}} \cr & \left( {\root 3 \of {1 + a} - 1} \right){x^2} + \left( {\sqrt {1 + a} - 1} \right)x + \left( {\root 6 \of {1 + a} - 1} \right) = 0 \cr & \left( {{{\left( {1 + a} \right)}^{\frac{1}{3}}} - 1} \right){x^2} + \left( {{{\left( {1 + a} \right)}^{\frac{1}{2}}} - 1} \right)x + \left( {{{\left( {1 + a} \right)}^{\frac{1}{6}}} - 1} \right) = 0.....\left( 1 \right) \cr & \mathop {\lim }\limits_{a \to {0^ + }} \alpha \left( a \right),\,\mathop {\lim }\limits_{a \to {0^ + }} \beta \left( a \right){\text{ and }}\alpha \left( a \right),\,\beta \left( a \right){\text{ are the roots}} \cr & {\text{Put}}\,1 + a = y,{\text{ equation }}\left( 1 \right){\text{ can be re - written as:}} \cr & \left( {{y^{\frac{1}{3}}} - 1} \right){x^2} + \left( {{y^{\frac{1}{2}}} - 1} \right)x + \left( {{y^{\frac{1}{6}}} - 1} \right) = 0 \cr & \Rightarrow \frac{{\left( {{y^{\frac{1}{3}}} - 1} \right){x^2}}}{{\left( {y - 1} \right)}} + \frac{{\left( {{y^{\frac{1}{2}}} - 1} \right)x}}{{\left( {y - 1} \right)}} + \frac{{\left( {{y^{\frac{1}{6}}} - 1} \right)}}{{\left( {y - 1} \right)}} = 0 \cr & {\text{Taking }}y \to 1,{\text{ as }}a \to 0 \cr & \Rightarrow \frac{{\left( {{y^{\frac{1}{3}}} - 1} \right){x^2}}}{{\left( {{y^{\frac{1}{3}}} - 1} \right)\left( {{y^{\frac{2}{3}}} + 1 + 2{y^{\frac{1}{3}}}} \right)}} + \frac{{\left( {{y^{\frac{1}{2}}} - 1} \right)x}}{{\left( {{y^{\frac{1}{2}}} - 1} \right)\left( {{y^{\frac{1}{2}}} + 1} \right)}} + \frac{{\left( {{y^{\frac{1}{6}}} - 1} \right){x^2}}}{{\left( {{y^{\frac{1}{6}}} - 1} \right)\left( {{y^{\frac{5}{6}}} + 2{y^{\frac{1}{6}}} + 2{y^{\frac{3}{6}}} + 1} \right)}} = 0 \cr & {\text{As putting }}y \to 1,{\text{ we get}} \cr & \Rightarrow \frac{{{x^2}}}{3} + \frac{x}{2} + \frac{1}{6} = 0 \cr & \Rightarrow 2{x^2} + 3x + 1 = 0 \cr & {\text{we get, }}x = - 1,\, - \frac{1}{2} \cr & {\text{Hence, option B is correct}}{\text{.}} \cr} $$

The given function can be restated as
\[f\left( x \right) = \left\{ \begin{array}{l} \frac{{\sin \left[ x \right]}}{{\left[ x \right]}},\,\,\,\,{\rm{if}}\,x \in \left( { - \,\infty ,0} \right) \cup \left[ {1,\,\infty } \right]\\ 0\,\,\,\,\,,\,\,\,\,\,\,\,{\rm{if}}\,x \in \left[ {0,\,1} \right) \end{array} \right.\]
$$\eqalign{ & \therefore \mathop {\lim }\limits_{x\, \to \,{0^ - }} f\left( x \right) = \mathop {\lim }\limits_{h\,\, \to \,0} \frac{{\sin \left[ { - h} \right]}}{{\left[ { - h} \right]}} \cr & = \mathop {\lim }\limits_{x\, \to \,{0^ - }} \frac{{\sin \left( { - 1} \right)}}{{\left( { - 1} \right)}} = \sin \,1 \cr & {\text{And}}\mathop {\lim }\limits_{x\, \to \,{0^ + }} f\left( x \right) = \mathop {\lim }\limits_{h\, \to \,0} 0 = 0 \cr & \because \mathop {\lim }\limits_{x\, \to \,{0^ - }} f\left( x \right) \ne \mathop {\lim }\limits_{x\, \to \,{0^ + }} f\left( x \right)\,\,\,{\text{as}}\,\,\sin \,1 \ne 0 \cr & \therefore \mathop {\lim }\limits_{x\, \to \,0} f\left( x \right)\,\,{\text{does not exist}}{\text{.}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \frac{{{x^2}}}{4}\log \left( {1 + 3x} \right)}} \cr & = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {{4^x} - 1} \right)}^3}}}{{{x^3}}}.\frac{{{{\left( {\frac{x}{2}} \right)}^2}}}{{\sin \frac{{{x^2}}}{4}}}.\frac{{3x}}{{\log \left( {1 + 3x} \right)}}.\frac{4}{3} \cr & = \frac{4}{3}{\left( {{{\log }_e}4} \right)^3}.1.{\log _e}\left( e \right) \cr & = \frac{4}{3}{\left( {{{\log }_e}4} \right)^3} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x\, \to \,0} \frac{{x\,\tan \,2x - 2x\,\tan \,x}}{{{{\left( {1 - \cos \,2x} \right)}^2}}} \cr & = \mathop {\lim }\limits_{x\, \to \,0} \frac{{x\left\{ {2x + \frac{{8{x^3}}}{3} + \frac{{64{x^5}}}{{15}} + ...} \right\} - 2x\left\{ {x + \frac{{{x^3}}}{3} + \frac{{2{x^5}}}{{15}} + ...} \right\}}}{{4{{\sin }^4}x}} \cr & = \mathop {\lim }\limits_{x\, \to \,0} \frac{{{x^4}\left\{ {\frac{8}{3} - \frac{2}{3} + {\text{ terms containing higher positive powers of }}x} \right\}}}{{4{{\sin }^4}x}} \cr & = \frac{1}{4}.2 \cr & = \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{We have, }}\frac{d}{{dx}}\left\{ {{{\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)}^2}} \right\} \cr & = \frac{d}{{dx}}\left\{ {x + \frac{1}{x} + 2} \right\} \cr & = \frac{d}{{dx}}\left( x \right) + \frac{d}{{dx}}\left( {{x^{ - 1}}} \right) + \frac{d}{{dx}}\left( 2 \right) \cr & = 1 + \left( { - 1} \right){x^{ - 2}} + 0 \cr & = 1 - \frac{1}{{{x^2}}} \cr} $$

$$\eqalign{ & \mathop {\lim }\limits_{x \to 5} \frac{{{{\left( {f\left( x \right)} \right)}^2} - 9}}{{\sqrt {\left| {x - 5} \right|} }} = 0 \cr & \mathop {\lim }\limits_{x \to 5} \left[ {{{\left( {f\left( x \right)} \right)}^2} - 9} \right] = 0\,\,\,\,\, \Rightarrow \mathop {\lim }\limits_{x \to 5} f\left( x \right) = 3 \cr} $$