Limits and Cont...

lf $$f(x)=2x-3,a=2,l=1$$ and $$\epsilon =0.001$$ then $$\delta>0$$ satisfying$$ 0<|x-a|<\delta, \ \ |f(x)-l|<\epsilon$$, is:

$$\displaystyle \lim_{x\rightarrow 0}\frac{e^{x}-e^{\sin x}}{2(x-\sin x)}=$$

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{x\tan 2x-2x\tan x}{(1-\cos 2x)^{2}}$$=

lf $$ \displaystyle \lim _{ x\rightarrow 0 } \left(\displaystyle  \frac { \cos  4x+a\cos  2x+b }{ x^{ 4 } }  \right) $$ is finite then the value of $$a,b$$ respectively are

$$\displaystyle \lim_{x\rightarrow 1}(1-x)\tan(\displaystyle \frac{\pi x}{2})=$$

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{1-\cos^{3}x}{x\sin 2x}$$=

$$\displaystyle \lim_{x\rightarrow \dfrac{\pi }{2}}\displaystyle \dfrac{(\dfrac{\pi}{2}-x)\sec x}{cosecx}$$=

$$\displaystyle \lim_{x\rightarrow 0}\displaystyle \frac{(1-\cos 2x)\sin 5x}{x^{2}\sin 3x}=$$

Solve : $$\displaystyle \lim_{x\rightarrow 0}\displaystyle \dfrac{\sin x\sin\left(\dfrac{\pi}{3}+x\right)\sin\left(\dfrac{\pi}{3}-x\right)}{x}$$

The value of $$\displaystyle \lim _{ \alpha \rightarrow \beta  }{ \left[ \frac { \sin ^{ 2 }{ \alpha  } -\sin ^{ 2 }{ \beta  }  }{ { \alpha  }^{ 2 }-{ \beta  }^{ 2 } }  \right]  } $$ is: