Question

$Let space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell table row cell 2 straight x end cell end table if space straight x less than 2 end cell row cell table row cell 2 space if space straight x equals 2 end cell row cell straight x squared space if space straight x greater than 2 end cell end table end cell end table close$
Show that 2 is a removable discontinuity of f.

Solution

Here space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell table row cell 2 straight x end cell end table if space straight x less than 2 end cell row cell table row cell 2 space if space straight x equals 2 end cell row cell straight x squared space if space straight x greater than 2 end cell end table end cell end table close space Lt with straight x rightwards arrow 2 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of minus below 2 straight x space space space space space space space space space space space space space space space space left square bracket Put space straight x equals 2 minus straight h comma space straight h greater than 0 space si space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 2 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 equals Lt with straight h rightwards arrow 0 below open curly brackets 2 left parenthesis 2 minus straight h right parenthesis close curly brackets equals Lt with straight h rightwards arrow 0 below left parenthesis 4 minus 2 straight h right parenthesis equals 4 minus 0 equals 4 space Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of plus below straight x squared space space space space space space space space space space space space space space space space left square bracket Put space straight x equals 2 plus straight h comma space straight h greater than 0 space si space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 2 to the power of plus right square bracket 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 left parenthesis 2 plus straight h right parenthesis squared equals Lt with straight h rightwards arrow 0 below left parenthesis 4 plus 4 straight h plus straight h squared right parenthesis equals 4 plus 0 plus 0 equals 4 space Lt with straight x rightwards arrow 2 to the power of minus below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 2 to the power of plus below straight f left parenthesis straight x right parenthesis equals 4 space space space space space space space rightwards double arrow space Lt with straight x rightwards arrow 2 below straight f left parenthesis straight x right parenthesis equals 4 But space space space space straight f left parenthesis 2 right parenthesis equals 2 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 rightwards double arrow space Lt with straight x rightwards arrow 2 below straight f left parenthesis straight x right parenthesis not equal to straight f left parenthesis 2 right parenthesis

∴ f is discontinuous at x = 2 and this discontinuity is removable discontinuity due to the fact that if we take f (2) = 4, f becomes continuous.

For Daily Free Study Material Join wiredfaculty

Continuity And Differentiability

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location