Prove that the function f(x) = 5 x – 3 is continuous at x = 0,
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    Continuity and Differentiability
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    Examine the continuity of the function f(x) = 2x2 –1 at.x = 3

    Here
          straight f left parenthesis straight x right parenthesis equals 2 straight x squared minus 1 Lt with straight x rightwards arrow 3 below space straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow 3 below space left parenthesis 2 straight x squared minus 1 right parenthesis equals 2 left parenthesis 3 right parenthesis squared minus 1 equals 2 left parenthesis 9 right parenthesis minus 1 equals 18 minus 1 equals 17 Now space straight f space is space defined space at space straight x equals 3 and space space space space straight f left parenthesis straight x right parenthesis equals 2 left parenthesis 3 right parenthesis squared minus 1 equals 2 left parenthesis 9 right parenthesis minus 1 equals 18 minus 1 equals 17 therefore Lt with straight x rightwards arrow 3 below space space space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 3 right parenthesis equals 17 therefore space straight f space is space continous space at space straight x equals 3.

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    Answer
    Examine the following functions for continuity :
    f(x)=x-5

    Here f(x) = x – 5
    Function f is defined for all real numbers.
    Let c be any real number.
    ∴ f(c) = c – 5
    Also space Lt with straight x rightwards arrow straight c below straight f left parenthesis straight x right parenthesis equals Lt with straight x rightwards arrow straight c below left parenthesis straight x minus 5 right parenthesis equals straight c minus 5 therefore space Lt with straight x rightwards arrow straight c below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight c right parenthesis
    ∴ f is continuous at x = c
    But c is any real number.
    ∴ f is continuous at every real number.

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    Prove that the function f(x) = 5 x – 3 is continuous at x = 0, at x  = – 3 and at x = 5.

    Here space straight f left parenthesis straight x right parenthesis equals 5 straight x minus 3 left parenthesis straight i right parenthesis space Lt with straight x rightwards arrow 0 below space straight f left parenthesis straight x right parenthesis equals stack space Lt with straight x rightwards arrow 0 below space left parenthesis 5 straight x minus 3 right parenthesis equals 5 left parenthesis 0 right parenthesis minus 3 equals 0 minus 3 equals negative 3 Now space space straight f space is space defined space at space straight x equals 0 and space space straight f left parenthesis 0 right parenthesis equals 5 left parenthesis 0 right parenthesis minus 3 equals 0 minus 3 equals negative 3 therefore space Lt with straight x rightwards arrow 0 below space straight f left parenthesis straight x right parenthesis equals space straight f left parenthesis 0 right parenthesis equals negative 3 therefore space straight f space is space continous space at space straight x equals 0.

    left parenthesis ii right parenthesis space stack Lt space with straight x rightwards arrow negative 3 below space straight f left parenthesis straight x right parenthesis equals stack space Lt with straight x rightwards arrow negative 3 below space left parenthesis 5 straight x minus 3 right parenthesis equals 5 left parenthesis negative 3 right parenthesis minus 3 equals negative 15 minus 3 equals negative 18 Now space straight f space is space defined space at space straight x equals negative 3 and space space straight f left parenthesis negative 3 right parenthesis equals 5 left parenthesis negative 3 right parenthesis minus 3 equals negative 15 minus 3 equals negative 18 therefore stack space Lt space with straight x rightwards arrow negative 3 below space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis negative 3 right parenthesis equals negative 18 therefore space straight f space is space continous space at space straight x equals negative 3.

    left parenthesis iii right parenthesis stack space Lt space with straight x rightwards arrow 5 below straight f left parenthesis straight x right parenthesis equals stack space Lt with straight x rightwards arrow 5 below space left parenthesis 5 straight x minus 3 right parenthesis equals 5 left parenthesis 5 right parenthesis minus 3 equals 25 minus 3 equals 22 Now space straight f space is space defined space at space straight x equals 5 and space space straight f left parenthesis 5 right parenthesis equals 5 left parenthesis 5 right parenthesis minus 3 equals 25 minus 3 equals 22 therefore Lt with straight x rightwards arrow 5 below space straight f left parenthesis straight x right parenthesis equals straight f left parenthesis 5 right parenthesis equals 22 therefore straight f space is space continous space at space straight x equals 5.
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    Prove that the function f(x) = xn is continuous at x = n, where n is a positive integer.
    Syntax error from line 1 column 421 to line 1 column 428.
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    Answer
    Examine the following functions for continuity :
    straight f left parenthesis straight x right parenthesis equals fraction numerator 1 over denominator straight x minus 5 end fraction
    Here space straight f left parenthesis straight x right parenthesis equals fraction numerator 1 over denominator straight x minus 5 end fraction
    For f to be defined,
    x – 5 ≠ 0 i.e. x ≠ 5
    ∴Df = Set of real number except 5 = R - { 5}
    Let c ≠ 5 be any real number.
    Also space Lt with straight x rightwards arrow straight c below straight f left parenthesis straight x right parenthesis minus Lt with straight x rightwards arrow straight c below open parentheses fraction numerator 1 over denominator straight x minus 5 end fraction close parentheses equals fraction numerator 1 over denominator straight c minus 5 end fraction therefore space Lt with straight x rightwards arrow straight c below straight f left parenthesis straight x right parenthesis equals straight f left parenthesis straight c right parenthesis
    ∴ f is continuous at x = c.
    But c ≠ 5 is any real number
    ∴ f is continuous at every real number c ∈ D
    ∴ f is continuous function.
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