#LIMITS IN CALCULUS BC HOW TO#
#LIMITS IN CALCULUS BC CODE#

To clarify this a little more, let’s look at a similar problem suggested by Sondra Edwards on the Facebook site: Consider this similar function: So we need to find the value of f as its argument approaches.

But, notice that as x approaches 0 from both sides, the limit 2 is approached from the left (from below). Since f is not continuous at 2, the theorem cannot be used. Given the graph of a function f, shown at the left, what is ? In this form the limit is obviously 3.Įxample 2: The second example is also based on a graph. Although we cannot be certain, it appears that. Therefore, as f approaches 1 from the leftĪnother approach is to try to write the equation of f. However, on closer examination, we see that is always less than 1, so is approaching 1 from the left (or from below). Īt first glance it appears that as x approaches zero, approaches 1 and the limit does not exist since f is not continuous at 1, so the theorem cannot be used. Students were given the graph at the right and asked to find. Let’s look at some.Įxample 1:The first example is from the 2016 BC International exam, question 88. This does not mean that the limits necessarily do not exist, rather that we need to find some other way of determining them. The problem is that in the examples one or the other of the hypotheses (continuity or the existence of ) is not met. The theorem that we would like to apply in these cases is this: Pastebin / JSFIDDLE.Recently, a number of questions about the limit of composite functions have been discussed on the AP Calculus Community bulletin board and also on the AP Calc TEACHERS – AB/BC Facebook page.

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#LIMITS IN CALCULUS BC HOW TO#

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#LIMITS IN CALCULUS BC CODE#

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Limits in calculus bc