1.

सिद्ध कीजिए कि `lim_(n to oo) n/2r^(2)sin ""(2pi)/(n)=pir^(2)`

Answer» `underset(nto oo)limn/2r^(2)sin ""(2pi)/(n)`
यदि `n=oo` तब फलन का रूप `ooxx0` हो जाता है।
`thereforeunderset(xto oo)limn/2r^(2)sin ""(2pi)/(n)=underset(x to oo)lim (r^(2)sin ""(2pi)/(n))/(2//n)`
`=underset(x to oo)lim (pi r^(2)sin""(2pi)/(n))/((2pi)/(n))=pir^(2)underset( ntooo)lim (sin ""(2pi)/(n))/((2pi)/(n))`
माना `(2pi)/(n)=theta` यदि `n=oo,` तब `theta=0`
`therefore n to ooimpliesthetato 0`
`therefore pir^(2)underset(n to oo)lim(sin (2pi//n))/(2pi//n)=pir^(2)underset(theta to 0)lim (sin theta)/(theta)`
`=pi r^(2)xx`
`=pir^(2)" "(इतिसिद्धम) " "(becauseunderset(thetato 0)lim(sin theta)/(theta)=1)`


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