Limits and Cont...

If $\displaystyle\lim_{x\rightarrow a}{(f(x)+g(x))}=2$ and $\displaystyle\lim_{x\rightarrow a}{(f(x)-g(x))}=1$, 

The function f is continuous on the closed interval $[1, 5]$ and values of the function are shown in the table above. If the values in the table are used to calculate a trapezoidal sum, the approximate value of $\int_{1}^{5}f(x)dx$ is

The value of $\displaystyle \underset { n\rightarrow \infty  }{ lim } \left( \frac { 1 }{ n+1 } +\frac { 1 }{ n+2 } +...+\frac { 1 }{ 6n }  \right)$ is

A point where function $f(x)$ is not continuous where $f(x)=\left[ \sin { \left[ x \right]  }  \right]$ in $\left( 0,2\pi  \right)$; is ($\left[ \ast  \right]$ denotes greatest integer $\le x$)

The value of $\lim _{ x\rightarrow 0 }{ \left( { \left( \sin { x }  \right)  }^{ 1/x }+{ \left( 1+x \right)  }^{ \sin { x }  } \right)  }$ whre $x> 0$ is

$\underset { n \rightarrow \infty }{ Lt } \sum _{ r=2n+1\quad  }^{ 3n } \dfrac {n}{r^2 - n^2}$ is equal to :

The value of $\displaystyle \lim_{x \rightarrow 1^{-}}\dfrac {1 - \sqrt {x}}{(\cos^{-1}x)^{2}}$

$\displaystyle \lim_{I\rightarrow \left (\dfrac {\pi}{2}\right )} = \int_{0}^{t}\tan \theta \sqrt {\cos \theta} ln (\cos \theta) d\theta$ is equal to

If $f '$ (0) = 0 and f(x) is a differentiable and increasing function,then lim $x \rightarrow 0$  $\frac {x.f ' (x^2)}{f ' (x)}$

$\quad \lim _{ n\rightarrow \infty  }{ \cfrac { 1 }{ n } \sum _{ r=1 }^{ 2n }{ \cfrac { r }{ \sqrt { { n }^{ 2 }+{ r }^{ 2 } }  }  }  }$ equal to: