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Continuity And Differentiability

Question
CBSEENMA12034455
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Test the continuity of the function f(x) at the originstraight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator open vertical bar straight x close vertical bar over denominator straight x end fraction comma space space space space straight x not equal to 0 end cell row cell space space space 0 space space space space comma space space space space straight x equals 0 end cell end table close
then show that f(x) is discontinuous at x = 0.

Solution
Here space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator open vertical bar straight x close vertical bar over denominator straight x end fraction comma space space space space straight x not equal to 0 end cell row cell space space space 0 space space space space comma space space space space straight x equals 0 end cell end table close space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator open vertical bar straight x close vertical bar over denominator straight x end fraction equals space Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar 0 minus straight h close vertical bar over denominator 0 minus straight h end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket Put space straight x equals 0 minus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 to the power of minus right square bracket space space space space space space space space space space space space space space space space space equals space Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar negative straight h close vertical bar over denominator negative straight h end fraction equals space Lt with straight h rightwards arrow 0 below fraction numerator straight h over denominator negative straight h end fraction equals Lt with straight h rightwards arrow 0 below left parenthesis negative 1 right parenthesis equals negative 1 space Lt with straight x rightwards arrow 0 to the power of plus below space straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator open vertical bar straight x close vertical bar over denominator straight x end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar 0 plus straight h close vertical bar over denominator 0 plus straight h end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket Put space straight x equals 0 plus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 to the power of plus right square bracket space space space space space space space space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar straight h close vertical bar over denominator straight h end fraction equals space Lt with straight h rightwards arrow 0 below straight h over straight h equals 1 therefore space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis not equal to space Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis rightwards double arrow Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis space does space not space exist space space space space space space space space space space space space space rightwards double arrow space straight f space is space continous space at space straight x equals 0.

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