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Continuity And Differentiability

Question
CBSEENMA12034489
Wired Faculty App

Show that the function
straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator straight e to the power of 1 over straight x end exponent minus 1 over denominator straight e to the power of begin display style 1 over straight x plus 1 end style end exponent end fraction comma space straight x not equal to 0 end cell row cell space space space space space space 0 space space space space space space space comma space straight x equals 0 end cell end table close
is discontinous at x=0.

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 straight e to the power of 1 over straight x end exponent minus 1 over denominator straight e to the power of begin display style 1 over straight x plus 1 end style end exponent end fraction comma space straight x not equal to 0 end cell row cell space space space space space space 0 space space space space space space space comma space straight x equals 0 end cell end table close 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 minus below fraction numerator straight e to the power of 1 over straight x end exponent minus 1 over denominator straight e to the power of 1 over straight x end exponent plus 1 end fraction space space space space 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 h rightwards arrow 0 below fraction numerator straight e to the power of begin display style 1 over straight h end style end exponent minus 1 over denominator straight e to the power of 1 over straight h end exponent plus 1 end fraction equals fraction numerator 0 minus 1 over denominator 0 plus 1 end fraction space space space space space space space space space space space space space space space space space space space open square brackets because straight e to the power of negative 1 over straight h rightwards arrow 0 space as space straight h rightwards arrow 0 end exponent close square brackets space space space space space space space space space space space space space space space equals negative 1
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 to the power of plus below fraction numerator straight e to the power of begin display style 1 over straight x end style end exponent minus 1 over denominator straight e to the power of 1 over straight x end exponent plus 1 end fraction space space space space space space space space 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 space space space space space space space space space equals Lt with straight h rightwards arrow 0 below fraction numerator straight e to the power of begin display style 1 over straight h end style end exponent minus 1 over denominator straight e to the power of 1 over straight h end exponent plus 1 end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator begin display style straight e to the power of begin display style 1 over straight h end style end exponent over straight e to the power of begin display style 1 over straight h end style end exponent end style minus begin display style 1 over straight e to the power of begin display style 1 over straight h end style end exponent end style over denominator straight e to the power of begin display style 1 over straight h end style end exponent over straight e to the power of begin display style 1 over straight h end style end exponent plus 1 over straight e to the power of begin display style 1 over straight h end style end exponent end fraction equals fraction numerator 1 minus straight e to the power of negative begin display style 1 over straight h end style end exponent over denominator 1 plus straight e to the power of negative 1 over straight h end exponent end fraction space space space space space space space space space space space space space space space space equals fraction numerator 1 minus 0 over denominator 1 plus 0 end fraction space space space space space space space space space space space space space space space space space space space open square brackets because straight e to the power of negative 1 over straight h end exponent rightwards arrow 0 space as space straight h rightwards arrow 0 close square brackets space space therefore space Lt with straight x rightwards arrow 0 to the power of minus below straight f left parenthesis straight x right parenthesis not equal to Lt with straight x rightwards arrow 0 to the power of plus below straight f left parenthesis straight x right parenthesis rightwards double arrow space Lt with straight x rightwards arrow 0 below straight f left parenthesis straight x right parenthesis space does space not space exist
⇒ f(x) is discontinuous at x = 0.

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