Question

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

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 sin space straight x over denominator straight x end fraction comma space straight x less than 0 end cell row cell space straight x plus 1 space comma space straight x greater or equal than 0 end cell end table close 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 space space space space space left square bracket Put space straight x equals 0 minus 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 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 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 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 space space space space space space space right square bracket Put space straight x equals 0 plus straight h comma space straight h rightwards arrow 0 right square bracket space space space space space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 below left parenthesis 0 plus straight h plus 1 right parenthesis equals Lt with straight x rightwards arrow 0 below left parenthesis straight h plus 1 right parenthesis equals 0 plus 1 equals 1

Also f(0) = value of (x +1) at x = 0
= 0 + 1 = 1
$therefore Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 0 right parenthesis$
∴ f (x) is continuous at x = 0.

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