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