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Continuity And Differentiability

Question
CBSEENMA12034442
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A function f is defined as
straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x squared minus straight x minus 6 over denominator straight x squared minus 2 straight x minus 3 end fraction comma space space straight x not equal to 3 end cell row cell space space space space space space space space space 5 over 3 space space space space space space space space comma space straight x equals 3 end cell end table close
Prove that f is discontinuous at x = 3. Can the definition of f at x = 3 be modified so as to make it continuous there?

Solution
Here space space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight x squared minus straight x minus 6 over denominator straight x squared minus 2 straight x minus 3 end fraction comma space space straight x not equal to 3 end cell row cell space space space space space 5 over 3 space space space space space space space space space space space comma space straight x equals 3 end cell end table close space Lt with straight x rightwards arrow 3 below space straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow 3 below fraction numerator straight x squared minus straight x minus 6 over denominator straight x squared minus 2 straight x minus 3 end fraction space equals space Lt with straight x rightwards arrow 3 below space fraction numerator left parenthesis straight x minus 3 right parenthesis left parenthesis straight x plus 2 right parenthesis over denominator left parenthesis straight x minus 3 right parenthesis left parenthesis straight x plus 1 right parenthesis end fraction equals space Lt with straight x rightwards arrow 3 below fraction numerator straight x plus 2 over denominator straight x plus 1 end fraction equals fraction numerator 3 plus 2 over denominator 3 plus 1 end fraction equals 5 over 4 Now space space straight f left parenthesis 3 right parenthesis space equals 5 over 3 therefore space Lt with straight x rightwards arrow 3 below space straight f left parenthesis straight x right parenthesis not equal to straight f left parenthesis 3 right parenthesis
⇒ f(x) is discontinuous at x = 3.
If f is modified as straight f left parenthesis straight x right parenthesis equals 5 over 4, when x = 3, then f will be continuous at x = 3.

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