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What is essential and removable discontinuity of f(x)=(x-5)/(x^2-25)?

Dont really get how there can be both on one equation. Just started learning discontinuity so any help is much appreciated.

6 Answers

  • 1 month ago
    Favorite Answer

    The denominator can be factored:

    .... x - 5


    (x + 5)(x - 5)

    You can't divide by zero, so there are two values that won't work for x.

    x ≠ -5 and x ≠ 5

    So you have discontinuities at both -5 and 5.

    But if you notice, you can cancel out x - 5 from the top and bottom and you would have:

    .. 1


    x + 5

    So x = -5 is the first discontinuity (an asymptote) and the other is a removable discontinuity.

    Look at the graph in the link below. The first discontinuity is obvious at x = -5 because the graph before it is dipping down to -infinity but on the other side it is at +infinity coming down.

    At x = 5, that's a removable discontinuity. If you took out that little hole right there, the graph would have continued.

  • ?
    Lv 7
    1 month ago





    limit f(x)=limit{1/(x+5)}=1/10


    [when x=/=5 but very close to 5, (x-5)

    can be cancelled first. Thus, x=5 is a

    removable discontinuity]


    limit f(x)=limit{1/(x+5)}= -infinity.


    limit f(x)=limit{1/(x+5)}= infinity.


    [ x=-5 is a permanent discontinuity]

  • Mewtwo
    Lv 5
    1 month ago

    Looking at your function, you see that the function is not defined at x = 5 or x = -5 since:

    f(x) = (x - 5) / (x² - 25) =  (x - 5) / [(x - 5)(x + 5)].

    We see that there is a factor of x - 5 in both the numerator and the denominator so that what is on the right above is really just 1 / (x + 5). Thus, we are able to "remove" the discontinuity at x = 5 since we no longer have this factor in the denominator but cannot remove the discontinuity at x = -5. Thus, at x = 5 there is a removable discontinuity while at x = -5 the discontinuity is essential. What this really means is that if we took the limit as x approaches 5 of the original function, we would get an actual real number, in this case 1/10. To remove the discontinuity in f, we would simply define that we want f(5) = 1/10. However, at x = -5, we have an issue since the limit does not exist. In fact, as x approaches -5 from the left, f(x) tends to -∞. As x approaches -5 from the right, we see that f(x) approaches ∞. Thus, not only does the limit not exist, this function blows up on both sides of the line x = -5. This means that there must be a vertical asymptote there. Anytime that a function exhibits vertical asymptote behavior in a neighborhood of some x-value, this function has a discontinuity there that cannot be removed. That is a geometrical interpretation of an essential discontinuity. The analytical approach is to simply show that the limit of the function does not exist at the point which guarantees an essential discontinuity as this is the definition thereof.

  • 1 month ago

    There are discontinuities any time the denominator is zero.  So in your function:

    f(x) = (x - 5) / (x² - 25)

    If we factor that we get:

    f(x) = (x - 5) / [(x + 5)(x - 5)]

    So there are two discontinuities in this fucntion: x = -5 and x = 5.

    But if you look again, there is a common factor (x - 5) that can cancel each other out:

    f(x) = 1 / (x + 5)

    So the x = -5 discontinuity is one that cannot be removed but the x = 5 was removable.

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  • ?
    Lv 7
    1 month ago

    f(x) = (x - 5)/(x^2 - 25)

    f(x) = (x + 5)^-1

  • ?
    Lv 7
    1 month ago

    Because x^2 - 25 = (x+5)(x - 5), the discontinuity at x = 5 can be removed by redefining f(x) as 1/(x + 5).  However, the discontinuity at x = -5 is "essential" and can't be removed by an algebra trick.

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