Limits and Cont...

If $$f\left( x \right) =\sqrt { 1-\sqrt { 1-{ x }^{ 2 } }  } $$, then $$f(x)$$ is

$$\mathop {\lim }\limits_{x \to \pi /2} \left[ {x\tan x - \left( {\frac{\pi }{2}} \right)\sec x} \right]$$ is equal to

$$\underset { x\rightarrow 0 }{ lim } \cfrac { 8 }{ { x }^{ 8 } } \left( 1-cos\cfrac { { x }^{ 2 } }{ 2 } -cos\cfrac { { x }^{ 2 } }{ 4 } +cos\cfrac { { x }^{ 2 } }{ 2 } .cos\cfrac { { x }^{ 2 } }{ 4 }  \right) =$$

The value of  $$\lim _ { x \rightarrow \dfrac { 1 } { 2 } } \dfrac { 2 \sin ^ { - 1 } x - \dfrac { \pi } { 2 } } { 1 - 2 x ^ { 2 } }$$  is equal to

$$f(x) = \log_{1 - 2x}(1 + 2x)$$    for $$ x \ne 0$$
          $$= k$$                              for $$x = 0$$
is continuous at $$x = 0$$, find $$k.$$

If $$ \alpha , \beta , \ in (-\frac{\pi}{2},0) $$ such that $$(sin \alpha +sin \beta ) +\frac{sin \alpha }{sin \beta} =0 $$ and $$(sin \alpha +sin \beta ) \frac{sin \alpha}{sin \beta }=-1 $$ and $$\lambda =\begin{matrix} lim \\ n\rightarrow \infty  \end{matrix}\frac { 1+(2sin\quad theta\quad ){  }^{ 2n } }{ (2sin\quad \quad beta\quad ){  }^{ 2n } } $$ then

$$\mathop {\lim }\limits_{x \to \infty } \left( {\dfrac{{{x^2}\sin \left( {\dfrac{1}{x}} \right) - x}}{{1 - \left| x \right|}}} \right) = $$

$$\begin{matrix} lim \\ x\rightarrow 0 \end{matrix}(cos\quad +\quad sinx{ ) }^{ 1/x }$$ is equal to

$$ \lim _{x \rightarrow a}\left(2-\frac{a}{x}\right)^{\tan \left(\frac{\pi x}{2 a}\right)} $$

$$\lim _{ x\rightarrow 0 }{ \frac { \sin { [\cos { x } ] }  }{ 1+[\cos { x } ] }  } $$ is (where [] is G.I.F)