Question

Determine if f is definend by
$straight f left parenthesis straight x 0 right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell xsin 1 over straight x comma space straight x not equal to 0 end cell row cell space space 0 space space space space comma space straight x equals 0 end cell end table close$

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Solution

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell xsin 1 over straight x comma space straight x not equal to 0 end cell row cell space space 0 space space space space comma space straight x equals 0 end cell end table close space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow 0 to the power of minus below straight x space sin 1 over straight x 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 space space space space space space equals Lt with straight x rightwards arrow 0 below left parenthesis negative straight h right parenthesis sin open parentheses fraction numerator 1 over denominator 0 minus straight h end fraction close parentheses equals space Lt with straight x rightwards arrow 0 below left parenthesis negative straight h right parenthesis sin open parentheses negative 1 over straight h close parentheses space space space space space space space space equals Lt with straight x rightwards arrow 0 below left parenthesis negative straight h right parenthesis open parentheses negative sin 1 over straight h close parentheses equals Lt with straight x rightwards arrow below space straight h space sin 1 over straight h equals 0 space space space space space space space space space space space space space space space space space space space space open square brackets table row cell because Lt with straight x rightwards arrow 0 below straight h equals 0 space and space sin 1 over straight h space is space bounded space as space open vertical bar sin 1 over straight h close vertical bar less or equal than 1 end cell row cell therefore Lt with straight x rightwards arrow 0 below straight h space sin 1 over straight h equals 0 space as space we space know space that end cell row cell because Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis space straight g left parenthesis straight x right parenthesis equals 0 space if space Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis equals 0 space and space straight g left parenthesis straight x right parenthesis space is space bounded end cell end table close square brackets

Agian 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 plus below straight x space sin 1 over straight x 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 equals Lt with straight x 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 x rightwards arrow 0 below straight h space sin 1 over straight h equals 0 space space left square bracket As space expalained space above right square bracket Also space straight f left parenthesis 0 right parenthesis equals 0 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 is continuous at x = 0.

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