What will happen if the Removable Discontinuity is the same as the Vertical Asymptote?

Example: f(x)=16-4x^2/x^2-4x+4

It is both x=2

Update:

Example: f(x)=(16-4x^2)/(x^2-4x+4)

It is both x=2

7 Answers

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  • Vaman
    Lv 7
    1 month ago

    f(x)=(16-4x^2)/(x^2-4x+4)= (4-2x)(4+2x)/(x-2)^2=4(2-x)(2+x)/(2-x)^2= 4(2+x)/(2-x)

    x=2 is the singular point which can not be removed.

  • 1 month ago

    f(x)=(16-4x^2)/(x^2-4x+4)

    =>

    f(x)=4(2+x)(2-x)/(2-x)^2

    =>

    f(x)=4(2+x)/(2-x), if x=/=2.

    So, if the vertical asymptote x=2 exists, the

    discontinuity at x=2 exists too & vice versa.

    It is obvious that if the discontinuity is removed--

    that means the curve is continuous at that point.

    Thus, no vertical asymptote exists there.

  • 1 month ago

    Hello I recently answered this problem in a test, however I only got 5 out of 10 points because my teacher said that there was a removable discontinuity even though I think that only a infinite discontinuity exist... I just want to know what is correct

  • 1 month ago

    It is not acceptable for someone studying calculus to have trouble with formatting fractions. Do try to focus.

    Generally, a removable discontinuity is called removable because, you can make the function continuous by adding just one point (it looks like a hole).

    So a vertical asymptote is not a removable discontinuity.

    (16-4x^2) / (x^2-4x+4) = (4+2x) / (x-2) goes to infinity as x->2 so it's a vertical asymptote.

     The definition may differ from textbook to textbook though.

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  • 1 month ago

    A simpler example would be the function f(x) = x/x

    I'd say it's undefined at x=0, but in the limit is equal to 1 at that point.

  • 1 month ago

    Except for x=0, you have given us the equation of a line.

    Let's simplify the function you provided. Division takes precedence.

    16 - (4x²/x²) - 4x + 4

    16 - 4 - 4x + 4

    -4x + 16

    There's a hole at x=0, but no asymptote.

    Perhaps you meant (16 - 4x²)/(x² - 4x + 4)?

    -4(x - 2)(x + 2)/[(x - 2)²]

    = -4(x + 2)/(x - 2)

    x = 2 is not in the domain, but that's okay because you still have another x - 2 in the denominator to create a vertical asymptote at x=2.

    (You also have a horizontal asymptote at y=-4)

  • rotchm
    Lv 7
    1 month ago

    It typically makes a hole. But not always... that's because...

    well, you need to learn to write correctly. Did you really mean 

    f(x)=16-4x^2/x^2-4x+4 ?  (no you didn't btw). 

    Correct that and we will explain further. 

    To you own reply. No, there are no removable discontinuity in your function. 

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