Question

Find all points of discontinuity of f where
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space straight x over denominator straight x end fraction comma space space if space straight x less than 0 end cell row cell space space straight x plus 1 comma space space space if space straight x greater or equal than 0 end cell end table close$

Solution

$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space straight x over denominator straight x end fraction comma space space if space straight x less than 0 end cell row cell space space straight x plus 1 comma space space space if space straight x greater or equal than 0 end cell end table close$
Function f is defined for all points of the real line.
Let c be any real number.
Three cases arise :
Case I: c < 0
$space Lt with straight x rightwards arrow straight c below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow straight c below fraction numerator sin space straight x over denominator straight x end fraction equals fraction numerator sin space straight c over denominator straight c end fraction space space space space space space space space straight f left parenthesis straight c right parenthesis equals fraction numerator sin space straight c over denominator straight c end fraction therefore space Lt with straight x rightwards arrow straight c below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight c right parenthesis therefore space straight f space is space continous space at space all space points space straight x less than 0.$

Case II:c > 1
$space Lt with straight x rightwards arrow straight c below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow straight c below left parenthesis straight x plus 1 right parenthesis equals straight c plus 1 space space space space space space straight f left parenthesis straight c right parenthesis equals straight c plus 1 therefore space Lt with straight x rightwards arrow straight c below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight c right parenthesis therefore space straight f space is space continous space at space all space points space straight x greater than 0$
Case III: c = 0
$space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator sin space straight x over denominator straight x end fraction 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 Lt with straight x rightwards arrow 0 below fraction numerator sin left parenthesis 0 minus straight h right parenthesis over denominator 0 minus straight h end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator sin left parenthesis negative straight h right parenthesis over denominator negative straight h end fraction space space space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 below fraction numerator negative sin space straight h over denominator negative straight h end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator sin space straight h over denominator straight h end fraction equals 1 therefore space Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of plus below left parenthesis straight x plus 1 right parenthesis equals 0 plus 1 equals 1 Also space straight f left parenthesis 0 right parenthesis equals 0 plus 1 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 equals Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 0 right parenthesis$
∴ f is continuous at x = 0
Thus f has no point of discontinuity.

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