Discuss the continuity at x = 0 of the function from Mathematic

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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

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

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

Here 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 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 Lt with straight x rightwards arrow 0 to the power of minus below space straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 0 to the power of minus below open vertical bar straight x close vertical bar cos 1 over straight x 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 equals Lt with straight x rightwards arrow 0 below open vertical bar 0 minus straight h close vertical bar cos fraction numerator 1 over denominator 0 minus straight h end fraction space space space space space space space space space equals Lt with straight h rightwards arrow 0 below straight h space cos 1 over straight h equals 0 space space space space space space space space space space space space space space space space space space space space space open square brackets table row cell because Lt with straight x rightwards arrow 0 below straight h equals 0 and space cos 1 over straight h is space bounded space function space as space open vertical bar cos space 1 over straight h close vertical bar less or equal than 1 end cell row cell therefore Lt with straight h rightwards arrow 0 below straight h space cos 1 over straight h equals 0 space as space we space know space that space Lt with straight x rightwards arrow straight a below straight f left parenthesis straight x right parenthesis straight g left parenthesis straight x right parenthesis equals 0 end cell row cell if space Lt with straight x rightwards arrow straight a below straight f left parenthesis straight x right parenthesis equals 0 space and space straight g left parenthesis straight x right parenthesis space is space bounded space in space deleted space straight n space straight b space straight d space of space 0. end cell end table close square brackets 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 open vertical bar straight x close vertical bar cos 1 over straight x 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 space space space space space space space space space space space left square bracket Puttting 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 and 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 equals Lt with straight h rightwards arrow 0 below open vertical bar 0 plus straight h close vertical bar cos space fraction numerator 1 over denominator 0 plus straight h end fraction equals Lt with straight h rightwards arrow 0 below straight h space cos 1 over straight h equals 0 space space left square bracket As space expalined space above right square bracket Also space straight f left parenthesis 0 right parenthesis equals 0 therefore 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 plus below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 0 right parenthesis

∴f (x) is continuous at x = 0.

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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.

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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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