Show that the functionis discontinous at x=0. from Mathematics

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Continuity and Differentiability

Long Answer Type

31. Discuss continuity of f(x) at x = 0, if

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator square root of 1 plus straight x minus end root square root of 1 minus straight x end root over denominator sin space straight x end fraction comma space if space space straight x not equal to 0 end cell row cell space space space space space space space space space space space space 1 space space space space space space space space space space space space space space space space space space comma space if space straight x equals 0 end cell end table close

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Short Answer Type

32.

Determine if f is defined by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell straight x squared space sin 1 over straight x comma space straight x not equal to 0 end cell row cell space space 0 space space space space space comma space straight x equals 0 end cell end table close$

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33.

Determine if f define by
$straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space straight x over denominator straight x end fraction plus space cos space straight x comma space straight x not equal to 0 end cell row cell space space space space space space space space space space 2 space space space space space comma space straight x equals 0 end cell end table close at space straight x equals 0$

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Long Answer Type

34.

Determine if f is definend by
$straight f left parenthesis straight x 0 right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell xsin 1 over straight x comma space straight x not equal to 0 end cell row cell space space 0 space space space space comma space straight x equals 0 end cell end table close$

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Short Answer Type

35. Find all points of continuity of f, where

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator sin space straight x over denominator straight x end fraction comma space straight x less than 0 end cell row cell space straight x plus 1 space comma space straight x greater or equal than 0 end cell end table close at space straight x equals 0

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36. Discuss the continuity of the function

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell fraction numerator 1 minus cos 4 straight x over denominator straight x squared end fraction comma space straight x not equal to 0 end cell row cell space space space space space space space 8 space space space space comma space straight x equals 0 end cell end table close at space straight x equals 0

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Long Answer Type

37.

If space straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell open vertical bar straight x close vertical bar sin 1 over straight x comma space if space straight x not equal to 0 end cell row cell space space space space space space space 0 comma space space space space space if space straight x equals 0 end cell end table close

then discuss continuity of f(x) at x = 0.

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Short Answer Type

38. Discuss the continuity at x = 0 of the function

straight f left parenthesis straight x right parenthesis equals open curly brackets table attributes columnalign left end attributes row cell open vertical bar straight x close vertical bar space cos 1 over straight x comma space straight x equals 0 end cell row cell space space space space space space 0 space space space space comma space straight x equals 0 end cell end table close

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39. 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.

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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40.

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.

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