1.

Examine the existence of the following limits:lim_(xto0) cosecx

Answer»

Solution :L.H.L.`=lim_(xto0-)` COSECX
`=lim_(hto0)` COSEC(-h)
`=lim_(hto0)` -1/sin h`=-infty`
R.H.L.`=lim_(hto0+)`cosecx
`=lim_(hto0)`cosec(h)=`infty`
As L.H.L.neR.H.L. the LIMIT does not exist.


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