Limits and Cont...

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

$$x$$	$$1$$	$$2$$	$$3$$	$$4$$	$$5$$
$$f(x)$$	$$4$$	$$3$$	$$7$$	$$1$$	$$3$$

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: