{"id":11854,"date":"2018-01-20T01:42:35","date_gmt":"2018-01-20T09:42:35","guid":{"rendered":"https:\/\/magoosh.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":"https:\/\/magoosh.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],"class_list":["post-11854","post","type-post","status-publish","format-standard","hentry","category-ap","tag-ap-calculus"],"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. 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