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Continuity And Differentiability

Question
CBSEENMA12034577
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Let f be defined by

straight f left parenthesis straight x right parenthesis open curly brackets table attributes columnalign left end attributes row cell space space space space space straight x squared over straight x space space space space space space space space space space space space comma space 0 space less or equal than straight x less or equal than 1 end cell row cell 2 straight x squared minus 3 straight x plus 3 over 2 comma space 1 less than straight x less or equal than 2 end cell end table close
Show that f is continous at x=1.

Solution
Here space straight f left parenthesis straight x right parenthesis open curly brackets table attributes columnalign left end attributes row cell space space space space space straight x squared over straight x space space space space space space space space space space space space comma space 0 space less or equal than straight x less or equal than 1 end cell row cell 2 straight x squared minus 3 straight x plus 3 over 2 comma space 1 less than straight x less or equal than 2 end cell end table close space Lt with straight x rightwards arrow 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow 1 to the power of plus below open parentheses straight x squared over 2 close parentheses space space space space space space space space left square bracket Put space straight x equals 1 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 1 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 Lt with straight h rightwards arrow 0 below fraction numerator left parenthesis 1 minus straight h right parenthesis squared over denominator 2 end fraction equals space Lt with straight h rightwards arrow 0 below fraction numerator straight h squared minus 2 straight h plus 1 over denominator 2 end fraction equals fraction numerator 0 minus 0 plus 1 over denominator 2 end fraction equals 1 half space Lt with straight x rightwards arrow 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of plus below open parentheses 2 straight x squared minus 3 straight x plus 3 over 2 close parentheses 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 space Put space straight x equals 1 plus straight h comma space straight h greater than 0 right square bracket 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 open curly brackets 2 left parenthesis 1 plus straight h right parenthesis squared minus 3 left parenthesis 1 plus straight h right parenthesis plus 3 over 2 close curly brackets 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 open curly brackets 2 straight h squared plus 4 straight h plus 2 minus 3 minus 3 straight h plus 3 over 2 close curly brackets equals 0 plus 0 plus 2 minus 3 minus 0 plus 3 over 2 equals 1 half space Also space straight f left parenthesis 1 right parenthesis equals Value space of space open parentheses straight x squared over 2 close parentheses space at space straight x equals 1 space space space space space space space space space space space space space space space space space equals 1 half therefore Lt with straight x rightwards arrow 1 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 1 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 1 right parenthesis
∴ f  is continuous at x = 1.

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