Limits and Cont...

Consider $$A=\begin{bmatrix} \cos { \theta  }  & \sin { \theta  }  \\ -\sin { \theta  }  & \cos { \theta  }  \end{bmatrix}$$, then the value of $$\lim_{n \rightarrow \infty} \dfrac{A^{n}}{n}$$ (where $$\theta \in R$$) is equal to 

If $$\displaystyle\lim_ { x\rightarrow \lambda  } { \left( 2-\dfrac { \lambda  }{ x }  \right)  }^{ \lambda tan\left( \dfrac { \pi x }{ 2\lambda  }  \right)  }=\frac { 1 }{ e } ,$$ then $$\lambda $$ is equal to-

$$\displaystyle \lim_{x \rightarrow 0}\dfrac {1}{x\sqrt {x}}\left(a\ arc\ tan \dfrac {\sqrt {x}}{a}-b\ arc\ \tan \dfrac {\sqrt {x}}{b}\right)$$ has the value equal to

For each $$t\in R$$, let [t] be the greatest integer less than or equal to t. Then, 
$$\underset { { x\rightarrow 0 }^{ + } }{ lim } x([\frac { 1 }{ x } ]+[\frac { 2 }{ x } ]+...+[\frac { 15 }{ x } ])$$

$$\displaystyle \lim _{ x\rightarrow 2 }{ \frac { \sqrt [ 3 ]{ 60+{ x }^{ 2 } } -4 }{ \sin { \left( x-2 \right)  }  }  } $$

The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ \csc^{ 4 }{ x } \int _{ 0 }^{ { x }^{ 2 } }{ \frac { ln\left( 1+4t \right)  }{ { t }^{ 2 }+1 }  } dt } $$ is 

The value of $$\displaystyle\lim_{n\rightarrow \infty}n(n\{ln (n)-ln (n+1)\}+1)$$ is?

$$\displaystyle \lim_{x\rightarrow \infty}{x^{2}\sin\left(\log_{e}\sqrt{\cos\dfrac{\pi}{x}}\right)}$$

$$\displaystyle \lim _{ x\rightarrow \frac { \pi  }{ 4 }  }{ { \left( \sin { 2x }  \right)  }^{ \sec ^{ 2 }{ 2x }  } }$$ is equal to 

$$\underset { x\rightarrow \frac { \pi  }{ 2 }  }{ lim } \frac { (1-sinx)({ 8x }^{ 2 }-{ \pi  }^{ 3 })cosx }{ { (\pi -2x) }^{ 4 } } $$