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Continuity And Differentiability

Question
CBSEENMA12034451
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Examine the  continuity of 
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 minus straight a close vertical bar over denominator straight x minus straight a end fraction space space space space space space straight x not equal to straight a end cell row cell space space space space space space space space 1 space space space space space space space space space space space straight x equals straight a end cell end table close space at space space straight x equals straight a

Solution
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 minus straight a close vertical bar over denominator straight x minus straight a end fraction space space space space space space straight x not equal to straight a end cell row cell space space space space space space space space 1 space space space space space space space space space space space straight x equals straight a end cell end table close space Lt with straight x rightwards arrow straight a to the power of minus below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow straight a to the power of minus below fraction numerator open vertical bar straight x minus straight a close vertical bar over denominator straight x minus straight a end fraction space left square bracket Put space straight x equals straight a 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 straight a to the power of minus right square bracket 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 straight a minus straight h minus straight a close vertical bar over denominator straight a minus straight h minus straight a end fraction 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 negative 1 space Lt with straight x rightwards arrow straight a to the power of plus below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow straight a to the power of plus below fraction numerator open vertical bar straight x minus straight a close vertical bar over denominator straight x minus straight a end fraction space left square bracket Put space straight x equals straight a plus straight h comma space straight h greater than space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow straight a right square bracket 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 straight a plus straight h minus straight a close vertical bar over denominator straight a plus straight h minus straight a end fraction equals space 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 Also space straight f left parenthesis straight a right parenthesis equals 1 therefore space Lt with straight x rightwards arrow straight a to the power of minus below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow straight a to the power of plus below space
∴ f  is continuous at x = a.

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