Question

Examine the continuity of f (x) at x = 0.
$If space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x minus open vertical bar straight x close vertical bar comma space straight x not equal to 0 end cell row cell space space space space space space 2 space space space comma space straight x equals 0 end cell end table close$

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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 minus open vertical bar straight x close vertical bar comma space straight x not equal to 0 end cell row cell space space space space space space 2 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 space straight f vertical line left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of minus below open curly brackets straight x minus open vertical bar straight x close vertical bar close curly brackets space equals Lt with straight x rightwards arrow 0 to the power of minus below open curly brackets left parenthesis 0 minus straight h right parenthesis minus open vertical bar 0 minus straight h close vertical bar close curly brackets 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 left square bracket space Put space straight x equals space 0 minus straight h comma space straight h greater than 0 space so space that space straight h 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 space space equals Lt with straight h rightwards arrow 0 below space open curly brackets negative straight h minus open vertical bar straight h close vertical bar close curly brackets space equals Lt with straight h rightwards arrow 0 below space open curly brackets negative straight h minus straight h close curly brackets equals negative 0 minus 0 equals 0 But space straight f left parenthesis 0 right parenthesis equals 2 therefore space Lt with straight x rightwards arrow 0 to the power of minus below space straight f left parenthesis straight x right parenthesis not equal to straight f left parenthesis 0 right parenthesis space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space straight f left parenthesis straight x right parenthesis space is space discontinous space at space straight x equals 0.

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