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લક્ષ

સાબિત કરો કે નીચેના લક્ષનું અસ્તિત્વ નથી :
$limit as straight x rightwards arrow 0 of space space fraction numerator open vertical bar space x space close vertical bar over denominator x end fraction$

અહીં,

$limit as straight x rightwards arrow 0 space straight _ of space space fraction numerator open vertical bar space straight x space close vertical bar over denominator straight x end fraction equals space limit as straight x rightwards arrow 0 space straight _ of space space fraction numerator negative straight x over denominator straight x end fraction space space space space open parentheses because space straight x space rightwards arrow space 0 space straight _ space space space rightwards double arrow space straight x space less than space 0 space space rightwards double arrow space open vertical bar space straight x space close vertical bar space equals space minus space straight x space close parentheses equals space limit as straight x rightwards arrow 0 space straight _ of space space space open parentheses negative 1 close parentheses equals space minus 1$

અને $limit as straight x rightwards arrow 0 subscript plus space of space fraction numerator open vertical bar space x space close vertical bar over denominator x end fraction$

$limit as straight x rightwards arrow 0 subscript plus of space space space straight x over straight x space space space space space open parentheses because space straight x space rightwards arrow space 0 subscript plus space space rightwards double arrow space straight x space greater than space 0 space space rightwards double arrow space open vertical bar space straight x space close vertical bar space equals space straight x close parentheses equals space limit as straight x rightwards arrow 0 subscript plus of space space space 1 equals space 1$

આમ,

$limit as straight x rightwards arrow 0 space straight _ of space space space fraction numerator open vertical bar space straight x space close vertical bar over denominator straight x end fraction space space not equal to space limit as straight x rightwards arrow 0 subscript plus of space space space fraction numerator open vertical bar space straight x space close vertical bar over denominator straight x end fraction therefore space space limit as straight x rightwards arrow 0 of space space space space fraction numerator open vertical bar space straight x space close vertical bar over denominator straight x end fraction space bold ન ું bold space bold અસ ્ ત િ ત ્ વ bold space bold નથ ી bold.$

સાબિત કરો કે નીચેનાં લક્ષનું અસ્તિત્વ નથી :
$limit as straight x rightwards arrow 0 of space space space fraction numerator open vertical bar space x space minus space 3 space close vertical bar over denominator x space minus space 3 end fraction$

અહીં, $limit as straight x rightwards arrow 3 of space space space fraction numerator open vertical bar space straight x space minus space 3 space close vertical bar over denominator straight x space minus space 3 end fraction$

$equals space limit as straight x rightwards arrow 3 of space space fraction numerator negative open parentheses straight x space minus space 3 close parentheses over denominator straight x space minus space 3 end fraction space space space space open parentheses because space straight x space rightwards arrow space 3 space straight _ space space space space rightwards double arrow space straight x space less than space 3 space space space rightwards double arrow space straight x space minus space 3 space less than space 0 space space rightwards double arrow space open vertical bar space straight x space minus space 3 space close vertical bar space equals space minus space open parentheses straight x space minus space 3 close parentheses close parentheses equals space limit as straight x rightwards arrow 3 space straight _ of space space space minus 1 equals space minus 1$

અને $limit as straight x rightwards arrow 3 subscript plus of space space fraction numerator open vertical bar x space minus space 3 close vertical bar over denominator x space minus space 3 end fraction space$

$equals space limit as straight x rightwards arrow 3 subscript plus of space space fraction numerator straight x space minus space 3 over denominator straight x space minus space 3 end fraction space space open parentheses because space straight x space rightwards arrow space 3 subscript plus space rightwards double arrow space straight x space greater than space 3 space space space rightwards double arrow space straight x space minus space 3 space greater than space 0 space space space rightwards double arrow space open vertical bar straight x space minus space 3 close vertical bar space space equals space straight x space minus space 3 close parentheses equals space limit as straight x rightwards arrow 3 subscript plus of space 1 space equals space 1$

$bold આમ bold comma bold space bold space bold space limit as straight x rightwards arrow 3 straight _ of space space fraction numerator open vertical bar straight x space minus space 3 close vertical bar over denominator straight x space minus space 3 end fraction space space not equal to space limit as straight x rightwards arrow 3 subscript plus of space space fraction numerator open vertical bar straight x space minus space 3 close vertical bar over denominator straight x space minus space 3 end fraction therefore space limit as straight x rightwards arrow 3 of space space fraction numerator open vertical bar straight x space minus space 3 close vertical bar over denominator straight x space minus space 3 end fraction space space bold space bold space bold ન ું bold space bold અસ ્ ત િ ત ્ વ bold space bold નથ ી bold.$

લક્ષના બીજગણિત ઉપયોગ કરી નીચેનો દાખલો સાબિત કરો :
$limit as straight x rightwards arrow 2 of space space x squared space equals space 4$

limit as straight x rightwards arrow 2 of space space x squared equals space open parentheses 2 close parentheses squared equals space 4

લક્ષના બીજગણિત ઉપયોગ કરી નીચેનો દાખલો સાબિત કરો :
$limit as straight x rightwards arrow 3 of space space x cubed space equals space open parentheses 3 close parentheses cubed equals space 27$

લક્ષના બીજગણિત ઉપયોગ કરી નીચેનો દાખલો સાબિત કરો :
$limit as straight x rightwards arrow 1 of space space open vertical bar space x space close vertical bar squared space equals space 1 space$

limit as straight x rightwards arrow 1 of space space open vertical bar space straight x space close vertical bar squared space equals space limit as straight x rightwards arrow 1 of space space straight x squared space equals space 1 squared space equals space 1

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