{"id":11854,"date":"2018-01-20t01:42:35","date_gmt":"2018-01-20t09:42:35","guid":{"rendered":"\/\/www.catharsisit.com\/hs\/?p=11854"},"modified":"2018-01-20t01:42:35","modified_gmt":"2018-01-20t09:42:35","slug":"ap-calculus-review-discontinuities","status":"publish","type":"post","link":"\/\/www.catharsisit.com\/hs\/ap\/ap-calculus-review-discontinuities\/","title":{"rendered":"ap calculus review: discontinuities"},"content":{"rendered":"

what’s a discontinuity? any point at which a function fails to be continuous<\/em><\/a> is called a discontinuity<\/strong>. in fact, there are various types of discontinuities, which we hope to explain in this review article.<\/p>\n

points of discontinuity<\/h2>\n

the definition of discontinuity is very simple. a function is discontinuous<\/strong> at a point x<\/em> = a<\/em> if the function is not<\/em> continuous at a<\/em>. <\/p>\n

so let’s begin by reviewing the definition of continuous<\/em>.<\/p>\n

a function f<\/em> is continuous<\/strong> at a point x<\/em> = a<\/em> if the following limit equation is true.<\/p>\n

\"the<\/p>\n

think of this equation as a set of three conditions.<\/p>\n

    \n
  1. the limit must exist. that is, \"limit for some finite number l<\/em>.<\/li>\n
  2. the function value must exist. in other words, f<\/em>(a<\/em>) exists.<\/li>\n
  3. the limit must agree with the function value. so, the number l<\/em> that you get by taking the limit should be the same value as f<\/em>(a<\/em>).<\/li>\n<\/ol>\n

    when one or more of these conditions fails, then the function has a discontinuity at x<\/em> = a<\/em>, by definition.<\/p>\n

    the first condition, that the limit must exist, is especially interesting. a limit may fail to exist for a variety of reasons. if the limit as x<\/em> → a<\/em> does not exist, then we can say that the function has a non-removable discontinuity<\/strong> at x<\/em> = a<\/em>. then, depending on how the limit failed to exist, we classify the point further as a jump<\/em>, infinite<\/em>, or infinite oscillation<\/em> discontinuity.<\/p>\n

    jump discontinuities<\/h2>\n

    one way in which a limit may fail to exist at a point x<\/em> = a<\/em> is if the left hand limit does not match the right hand limit.<\/p>\n

    in the graph shown below, there seems to be a “mismatch.” as x<\/em> approaches 1 from the left, that part of the graph seems to land on y<\/em> = -1. on the other hand, as x<\/em> approaches 1 from the right, the values of y<\/em> seem to get closer and closer to y<\/em> = 3.<\/p>\n

    \"piecewise
    a function f<\/em>(x<\/em>) with a jump discontinuity at x<\/em> = 1<\/figcaption><\/figure>\n

    in limit notation, we would write:<\/p>\n

    \"left<\/p>\n

    \"right<\/p>\n

    because the left and right limits do not agree, the limit of f<\/em>(x<\/em>) as x<\/em> → 1 does not exist<\/em>.<\/p>\n

    therefore, by definition, the function f<\/em> is discontinuous at x<\/em> = 1. this kind of discontinuity is known as a jump<\/strong> (for obvious reasons).<\/p>\n

    infinite discontinuities (vertical asymptotes)<\/h2>\n

    in some functions, the values of the function approach ∞ or -∞ as x<\/em> approaches some finite number a<\/em>. in this case, we say that the function has an infinite discontinuity<\/strong> or vertical asymptote<\/a><\/strong> at x<\/em> = a<\/em>.<\/p>\n

    you might think of an infinite discontinuity as an extreme case of jump discontinuity.<\/p>\n

    \"graph
    this function has an infinite discontinuity at x<\/em> = 3.<\/figcaption><\/figure>\n

    typically, you’ll find this behavior anywhere there is a division of the form nonzero over zero<\/em>. <\/p>\n

    for instance 3x<\/em>\/(x<\/em> + 12) has an infinite discontinuity at x<\/em> = -12, because that is where the denominator becomes 0 while the numerator is nonzero (3 × -12 = -36).<\/p>\n

    example<\/h3>\n

    locate and classify the discontinuities of f<\/em>(x<\/em>) = tan x<\/em> on the interval [-2π, 2π].<\/p>\n

    solution<\/h4>\n

    note that tan x<\/em> = (sin x<\/em>)\/(cos x<\/em>), and the denominator, cos x<\/em>, is equal to zero at all points of the form π\/2 + n<\/em>π. there are four x<\/em>-values within the interval [-2π, 2π] that qualify: x<\/em> = -3π\/2, -π\/2, π\/2 and 3π\/2. plugging each one into the numerator, sin x<\/em>, we verify that the top is nonzero when the bottom is zero. thus, f<\/em> has an infinite discontinuity at four points: x<\/em> = -3π\/2, -π\/2, π\/2 and 3π\/2 within the interval [-2π, 2π]<\/p>\n

    the graph can help us to visualize what’s going on near those point.<\/p>\n

    \"trigonometry
    graph of tan x<\/em>, showing infinite behavior at π\/2 + n<\/em>π.<\/figcaption><\/figure>\n

    infinite oscillation discontinuities<\/h2>\n

    most of the functions that you meet are fairly boring and predictable. even at a jump or infinite discontinuity, you can say something about how the values of the function behave. however if your function has an infinite oscillation discontinuity<\/strong> at a point x<\/em> = a<\/em>, then things get wild!<\/p>\n

    consider the graph of f<\/em>(x<\/em>) = sin(1\/x<\/em>).<\/p>\n

    \"graph
    graph of sin(1\/x<\/em>) on [-5, 5]<\/figcaption><\/figure>\n

    approaching the origin, the curve goes up and down more and more often. in fact as x<\/em> → 0, from either the left or right, we can see how the oscillations get completely out of hand! it’s almost as if the curve wants to make sure it hits every<\/em> point in the range between -1 and 1 at x<\/em> = 0.<\/p>\n

    no matter how far we zoom in on the graph, it just gets wilder and wilder.<\/p>\n

    \"graph
    graph of sin(1\/x<\/em>) on [-0.5, 0.5]<\/figcaption><\/figure>\n

    the reason for this strange behavior has to do with the fact that sin x<\/em> itself is periodic. then replacing x<\/em> by 1\/x<\/em> in the argument has the effect of taking all those infinitely many periodic waves of the sine function as x<\/em> → ∞ and squeezing<\/em> them next to the origin instead.<\/p>\n

    at any rate, since there is no single value of y<\/em> to which the curve seems to be heading as x<\/em> → 0, the limit does not exist at x<\/em> = 0.<\/p>\n

    therefore, f<\/em>(x<\/em>) = sin(1\/x<\/em>) has a discontinuity at x<\/em> = 0, of the infinite oscillation variety.<\/p>\n

    removable discontinuities<\/h2>\n

    the last category of discontinuity is different from the rest. in the previous cases, the limit did not exist. in other words, condition 1 of the definition of continuity failed. next we’ll discuss what happens if condition 1 holds (the limit exists), but either condition 2 or 3 fail.<\/p>\n

    a function f<\/em> has a removable discontinuity<\/strong> at x<\/em> = a<\/em> if the limit of f<\/em>(x<\/em>) as x<\/em> → a<\/em> exists, but either f<\/em>(a<\/em>) does not exist, or the value of f<\/em>(a<\/em>) is not equal to the limiting value.<\/p>\n

    if the limit exists, but f<\/em>(a<\/em>) does not, then we might visualize the graph of f<\/em> as having a “hole” at x<\/em> = a<\/em>. you might imagine what happens if you filled in that hole. if it really is a removable<\/em> discontinuity, then filling in the hole results in a continuous graph!<\/p>\n

    let’s take a look at the graph below.<\/p>\n

    \"graph<\/p>\n

    there is a hole at x<\/em> = -2. in fact, the graph would be continuous at that point if the hole at (-2, 2) were filled in. that’s the clue that we’re dealing with a removable <\/em>discontinuity at x<\/em> = -2.<\/p>\n

    there is one more case to consider. suppose both conditions 1 and 2 hold for a function at a given point, but condition 3 fails. that is, the limit exists, the function value exists, but they are different<\/em> values.<\/p>\n

    this situation happens in the graph shown below.<\/p>\n

    \"removable<\/p>\n

    focus at what happens near x<\/em> = 2. <\/p>\n

    there is a well-defined limit value. because the curve approaches y<\/em> = 3 from the left and the right, the limit is equal to 3. <\/p>\n

    but f<\/em>(2) = 1 (based on the location of the dot). since the value of f<\/em> is not the same as the limiting value here, we can say that f<\/em> has a removable discontinuity<\/em> at x<\/em> = 2.<\/p>\n

    example<\/h3>\n

    identify where the function has a removable discontinuity and determine the value of the function that would make it continuous at that point.<\/p>\n

    \"rational<\/p>\n

    solution<\/h4>\n

    look to the denominator. we know that f<\/em> will be undefined whenever the bottom of the fraction is equal to 0. that happens when 2x<\/em> – 4 = 0, or after solving for x<\/em>, we find x<\/em> = 2.<\/p>\n

    so we already know there is some kind of discontinuity at x<\/em> = 2 (because f<\/em>(2) does not exist — see condition 2 of the definition of continuity). but is it removable<\/em>? that takes finding a limit.<\/p>\n

    \"limit<\/p>\n

    because the limit value exists, we now know for sure that there is a removable discontinuity at x<\/em> = 2.<\/p>\n

    moreover, the limit itself tells us what the value should be to make f<\/em> continuous: -5\/2.<\/p>\n","protected":false},"excerpt":{"rendered":"

    what’s a discontinuity? any point at which a function fails to be continuous is called a discontinuity. in fact, there are various types of discontinuities, which we hope to explain in this review article. points of discontinuity the definition of discontinuity is very simple. a function is discontinuous at a point x = a if […]<\/p>\n","protected":false},"author":223,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[240],"tags":[241],"ppma_author":[24932],"acf":[],"yoast_head":"\nap calculus review: discontinuities - magoosh blog | high school<\/title>\n<meta name=\"description\" content=\"discontinuities are points at which a function fails to be continuous. there are various types of discontinuities. read on to find out more!\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" 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