Limits and Cont...

$$\underset{x \rightarrow \frac{\pi}{2}}{\lim} \dfrac{\cot x - \cos x}{\left(\dfrac{\pi}{2} -x \right)^3} = $$

$$\mathop {\lim }\limits_{x \to 0} \,\dfrac{1}{x}\,{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$$ is equal to

solve the limit 

Solve

$$\mathop {\lim }\limits_{x \to 0} \dfrac{1}{x}{\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$$ is equal to

$$\mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {2x - 3} \right)\left( {\sqrt x  - 1} \right)}}{{2{x^2} + x - 3}} = $$

If   $${z_r} = \cos \dfrac{{r\alpha }}{{{n^2}}} + i\sin \dfrac{{r\alpha }}{{{n^2}}}$$, where $$ r= 1, 2, 3, ....n$$, then $$\mathop {\lim }\limits_{n \to \infty } \left( {{z_1}.{z_2}.....{z_n}} \right)$$ is equal to 

If $$f(x)$$ is the integral of $$\dfrac{2\sin{x}-\sin{2x}}{x^{3}},\ x\neq 0$$. Find $$\lim _{ x\rightarrow 0 }{ f^{ ' }\left( x \right)  } $$, where $$f^{ ' }\left( x \right) =\dfrac{df{(x)}}{dx}$$

Find the value of limit $$\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 6 }  }{ \frac { 2\sin ^{ 2 }{ x } +\sin { x-1 }  }{ 2\sin ^{ 2 }{ x } -3\sin { x+1 }  } = } $$.