Limits and Cont...

The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ f\left( x \right) } $$ where $$f(x)=\dfrac {\cos (\sin x)-\cos x}{x^{4}}$$, is

$$\lim _{ x\rightarrow \infty  }{ \frac { 1 }{ x } \int _{ 0 }^{ x }{ \left( \sqrt { { t }^{ 2 }+5t } -t \right) dt }  }$$ 

The value of $$\displaystyle \lim _{ x\rightarrow a }{ \frac { \sqrt { x-b } -\sqrt { a-b }  }{ { x }^{ 2 }-{ a }^{ 2 } }  } \left( a>b \right) $$ is

$$\lim _ { x \rightarrow 0 } \dfrac { ( 27 + x ) ^ { 1 / 3 } - 3 } { 9 - ( 27 + x ) ^ { 2 / 3 } }$$  equals :

The value of $$\displaystyle \lim _{ x\rightarrow 0 }{ \frac { x }{ 5 }  } \left[ \frac { x }{ 2 }  \right] $$ (where $$[.]$$ denotes the greatest integer function) is

$$\underset { x\rightarrow 1 }{ Lim } \left[ { \left[ \frac { 4 }{ { x }^{ 2 }-{ x }^{ -1 } } -\frac { { 1-3x+x }^{ 2 } }{ { 1-x }^{ 3 } }  \right]  }^{ -1 }+\frac { 3\left( { x }^{ 4 }-1 \right)  }{ { x }^{ 3 }-{ x }^{ -1 } }  \right] =$$

$$\displaystyle \underset { x\rightarrow 0 }{ lim } \ \ \frac { ({ 1-\cos2x) }^{ 2 } }{ 2x \tan x-x \tan2x } $$ is :

The value of $$\displaystyle \lim_{x\rightarrow 0} \dfrac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}$$ is

$$\underset { x\rightarrow 0 }{ \lim } \dfrac { { 3 }^{ 2x }-{ 2 }^{ 3x } }{ x } $$ is equal to

Let $$f(x)=\dfrac{ax+b}{x+1},lim_{x\rightarrow 0} f(x)=2$$ and $$lim_{x\rightarrow \infty} f(x)=1$$ then $$f(-2)=$$