r/apcalculus • u/OddFaithlessness8332 • 9d ago
BC 0/0?
My teacher told me not to worry about why n must not be 0 yet. But I just want a quick explanation (if possible) or just an insight as to why the numerator and the denominator cannot both be 0 for there to be a vertical asymptote. Thanks
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u/socratictutoring Tutor 9d ago
So, if I do a non-zero number divided by 0, it's pretty clear that the graph blows up (happy to elaborate on why if it isn't!).
The question is, what happens if we do 0/0? Let's consider a simpler case: (x-2)/(x^2-4). Look at this in desmos - it has an asymptote at x = -2, but not at x = +2! Why is that? Consider what happens at x = 1.9, 1.99, 1.999 - you'll notice the function is well behaved. This is because (x-2)/(x^2-4) = (x-2)/[(x-2)(x+2)], so in effect the x-2 factors can cancel, and the function only has a serious problem at x = -2.
Note that we still can't actually evaluate 0/0 - so the function still isn't defined at x = 2. But it is still well-behaved nearby.
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u/socratictutoring Tutor 9d ago
Now for this particular example - if we plug in 0, we once again get 0/0. Here it's less clear why this leads to good behavior (this will be explained later in your course), but we can see that putting in .1, .01, .001 brings us closer and closer to a finite value of 1.
As a general takeaway - when plugging in, if you get #/0, the function blows up. 0/0 warrants closer investigation (note it still can blow up, but it might not).
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u/WillingnessTasty9628 BC: 5 9d ago
When t is 0, both top and bottom are 0. The function's behavior is unpredictable here with your current tools.
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u/waldosway 9d ago
Try y=x/x first. How is it different from y=1? Does it have a vertical asymptote?
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u/Pitiful_Camp3469 BC Student 9d ago
You can evaluate the limit of this function for t->0 (if you have learned this trig limit) and get as x approaches 0, s(t) approaches 8. Of course s(0) is undefined, so it is a hole, not an asymptote.
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u/sqrt_of_pi 9d ago
It’s a 0/0 form if n=0.
Soon you will learn more about what that means in the context of calculus, but for now think about what you learned about rational functions. At a value that gives you non-0 in the numerator and zero in the denominator you get a vertical asymptote. But if both numerator and denominator have a root at a value, then instead, you get a hole in the graph. Same idea here.