limits

[Maddox]

So we had our Calc final today and there was a question that threw me off.

(>_ is greater than or equal to)
x >_ 0, there is an asymptote at y = 2

(I eliminated the answers down to these 2)
a) y does not equal 2, x >_ 0
b) limit as x->infinity f(x) = 2

I put a because it had the x >_ 0, but I am not sure why b would be wrong.

[MyndFyre]

Hrm, this is tough.  I can see that A has all information given in the question.  That's probably a good, safe answer.

B, though --

The first example that comes to mind is f(x) = 1/x + 2.  By L'Hospital's Rule:

a(x) = 1/x + 2
a(x) = 1/x + 2x/x
a(x) = (1 + 2x) / x
L'Hospital's Rule: lim f(x)/g(x) = lim f'(x)/g'(x)
Let f(x) = 1+2x, g(x) = x
f'(x) = 2, g'(x) = 1
lim (1 + 2x) / x = lim 2/1 = 2

Intriguing.  If you find out, post!  I would think that it's some kind of technicality.  Example: because you know that it's asymptotic for x >= 0, consider:


Who knows what this function is?  Regardless:

lim f(x) as x->infinity = -infinity

[Adron]

Because x >= 0 is a closed interval end, the function can't approach 2 without also becoming 2 there. The asymptote must be at x -> infinity.

A is wrong. Example:

[Adron]

Btw, gogo post more math problems to think about, and give me more opportunities to run Maple?

[Maddox]

I got my final back yesterday, and yep, I was wrong. I still ended up setting the curve, though.

I actually had B marked, then changed it to A.

[Yoni]

MyndFyre's example is not horizontally asymptotic. To be horizontally asymptotic you have to satisfy lim x->infinity f(x) = Const or lim x->-infinity f(x) = Const. The former is the one that applies here because of the requirement that x >= 0. (Or - the question was badly phrased, which is most likely...)

MyndFyre: Looks kinda like 2-e^x.


Adron's counterexample for A was too complicated. Here is a simpler one:
y = 2 (for all x)

Asymptotic at y = 2, and equal to 2 pretty frequently (always).

[Adron]

Adron's counterexample for A was too complicated. Here is a simpler one:
y = 2 (for all x)

Asymptotic at y = 2, and equal to 2 pretty frequently (always).

I suppose that might work too.. It doesn't really fit "curve that approaches other curve", since it's equal to that other curve. But if it's all about having the same limit, it's perfect