Limits and Cont...

$$lim_{n\to \infty} \Sigma^n_{r=1} \dfrac{\pi}{n} sin(\dfrac{\pi r}{n})$$ is equal to

Evaluate : $$\displaystyle\lim _{ x\rightarrow 0 }{ \left( \dfrac { { e }^{ x\ell n\left( { 3 }^{ x }-1 \right)  }-\left( { 3 }^{ x }-1 \right) ^{ x }\sin { x }  }{ { e }^{ x\ell nx } }  \right)  } $$ is equal to 

$$\underset{x \rightarrow 0}{Lt} \dfrac{sin x - x + x^3 / 6}{x^5}$$ = 

$$\displaystyle \lim_{x\rightarrow0 }{\dfrac{(\cos\alpha)^{x}-(\sin\alpha)^{x}-\cos 2\alpha}{(x-4)}}, \alpha\in \left(0, \dfrac{\pi}{2}\right)$$ is equal to

evaluate$$ \underset { x\rightarrow 0 }{ lim } \frac { x-\int _{ 0 }^{ x }{ { cost }^{ 2 }dt }  }{ { x }^{ 3 }-6x } $$

$$\underset { x\rightarrow 0 }{ Lt } \cfrac {tanx-x}{x^2tanx}$$ equals:

$$\underset { x\rightarrow \pi/2 }{ lim } \left(\dfrac{cosec x-1}{cot^2x}\right)= $$

$$\displaystyle\lim_{x\rightarrow \infty}\left(\dfrac{x+1}{2x+1}\right)^{x^2}$$ equals?

$$\underset{x \rightarrow \infty}{lim} \dfrac{2 \tan^{-1} x}{\pi}$$ equals $$e^L$$ then $$L$$ is equal to

$$\displaystyle \lim _{ x\rightarrow x/2 } \dfrac { \left[ 1-\tan { \left( \dfrac { x }{ 2 }  \right)  }  \right] \left[ 1-\sin { x }  \right]  }{ \left[ 1+\tan { \left( \dfrac { x }{ 2 }  \right)  }  \right] \left[ \pi -2x \right] ^{ 3 } } $$ is