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Continuity And Differentiability

Question
CBSEENMA12034445
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Is straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator open square brackets straight x close square brackets minus 1 over denominator straight x minus 1 end fraction comma space straight x not equal to 1 end cell row cell space space space minus 1 space space space space space space comma space straight x equals 1 end cell end table close continous at x=1 ?

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 open square brackets straight x close square brackets minus 1 over denominator straight x minus 1 end fraction comma space straight x not equal to 1 end cell row cell space space space minus 1 space space space space space space comma space straight x equals 1 end cell end table close space Lt with straight x rightwards arrow 1 to the power of minus below space straight f left parenthesis straight x right parenthesis equals space Lt with straight x rightwards arrow 1 to the power of minus below space fraction numerator open square brackets straight x close square brackets minus 1 over denominator straight x minus 1 end fraction space left square bracket Put space straight x equals 1 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 1 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 equals space Lt with straight h rightwards arrow 0 below fraction numerator open square brackets 1 minus straight h close square brackets minus 1 over denominator 1 minus straight h minus 1 end fraction equals space Lt with straight h rightwards arrow 0 below fraction numerator 0 minus 1 over denominator 0 minus straight h end fraction equals space Lt with straight h rightwards arrow 0 below 1 over straight h comma
which does not exist
∴  f(x) is discontinuous at x = 1.

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