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Continuity And Differentiability

Question
CBSEENMA12034670

Examine the derivability of the following function:
straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x squared space sin 1 over straight x comma space straight x not equal to 0 end cell row cell space space space space space space space 0 space space space comma space space space space straight x equals 0 end cell end table close

Solution
Here space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x squared space sin 1 over straight x comma space straight x not equal to 0 end cell row cell space space space space space space space 0 space space space comma space space space space straight x equals 0 end cell end table close
straight L. straight H. straight D equals Lt with straight x rightwards arrow 0 below to the power of minus fraction numerator straight f left parenthesis straight x right parenthesis minus straight f left parenthesis 0 right parenthesis over denominator straight x minus 0 end fraction equals Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator straight x squared sin space begin display style 1 over straight x end style minus 0 over denominator straight x minus 0 end fraction equals Lt with straight x rightwards arrow 0 to the power of minus below open parentheses straight x space sin space 1 over straight x close parentheses
space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis 0 minus straight h right parenthesis sin open parentheses fraction numerator 1 over denominator 0 minus straight h end fraction close parentheses equals Lt with straight h rightwards arrow 0 below left parenthesis negative straight h right parenthesis sin open parentheses negative 1 over straight h close parentheses equals Lt with straight h rightwards arrow 0 below straight h space sin 1 over straight h
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 equals 0 space space space space space space space space space space space space space space space space space space space space space open square brackets because Lt with straight x rightwards arrow 0 below straight h equals 0 space and space open vertical bar sin 1 over straight h close vertical bar less or equal than 1 space in space deleted space nbd. space of space straight h equals 0 close square brackets
straight R. straight H. straight D equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator straight f left parenthesis straight x right parenthesis minus straight f left parenthesis 0 right parenthesis over denominator straight x minus 0 end fraction equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator straight x squared sin space begin display style 1 over straight x end style minus 0 over denominator straight x minus 0 end fraction equals Lt with straight x rightwards arrow 0 to the power of plus below open parentheses straight x space sin space 1 over straight x close parentheses
space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below left parenthesis 0 plus straight h right parenthesis sin open parentheses fraction numerator 1 over denominator 0 plus straight h end fraction close parentheses equals Lt with straight h rightwards arrow 0 below straight h space sin 1 over straight h
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 equals 0 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 space open square brackets as space explained space above close square brackets
therefore space straight L. straight H. straight D equals straight R. straight H. straight D equals 0
therefore space straight f left parenthesis straight x right parenthesis space is space derivalbe space at space straight x equals 0 space and space has space the space derivative space 0.

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