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Calculus: Easy limit problem - February 2nd 2011, 10:49 PM

If the limit of a function as x approaches 3 exists, would f(3) necessarily exist? Explain why.
   
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Re: Calculus: Easy limit problem - February 3rd 2011, 12:17 AM

This problem is so conceptual that it's impossible for it to be a respectable homework problem, so I feel no guilts about just explaining it.

Say you have the function f(x) = (x - 3) ^2 / (x - 3). If you graph this function, it'll look exactly like the function g(x) = x - 3, with the exception of a removable discontinuity at (3,0)

However, the left hand limit as x --> 3 is zero, and the right-hand limit as x --> 3 is also zero. Thus, the limit as x --> 3 is well-defined for f(x).

However again, f(3) is not defined.

Does this make sense?


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Re: Calculus: Easy limit problem - February 3rd 2011, 12:40 AM

Quote:
Originally Posted by Diaminohexanoate View Post
Say you have the function f(x) = (x - 3) ^2 / (x - 3). If you graph this function, it'll look exactly like the function g(x) = x - 3, with the exception of a removable discontinuity at (3,0)
If your f(x) function has a removable discontinuity, then would that mean the function would not have a limit?

Quote:
However, the left hand limit as x --> 3 is zero, and the right-hand limit as x --> 3 is also zero. Thus, the limit as x --> 3 is well-defined for f(x).
the graph I see, I'm not sure where you're getting zero.
   
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Re: Calculus: Easy limit problem - February 3rd 2011, 12:48 AM

Quote:
If your f(x) function has a removable discontinuity, then would that mean the function would not have a limit?
Nope. You're thinking of a jump discontinuity, which would occur in functions like y = int(x).

In fact, I'm almost certain that one criterion for a removable discontinuity is that the left-hand and right-hand limits converge to the same value at the discontinuity. That wouldn't occur in a jump discontinuity (going back to y = int(x), try to evaluate the left-hand and right-hand limits for x --> 1).

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the graph I see, I'm not sure where you're getting zero.
Why not? What graph do you see?


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Re: Calculus: Easy limit problem - February 3rd 2011, 12:53 AM

Okay, let me ask you a few questions. I would greatly appreciate your help.

Number 1, I'm confused on vertical asymptotes. My sister, who takes like Calc 4 or some shit, says that vertical asymptotes are not limits. If an equation has a vertical asymptote, that means the limit is undefined. Is that correct?

What about this? The problem asks "determine if the limit as x approaches 4 would always have the same value as f(4). Explain. Which function would you provide?
   
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Re: Calculus: Easy limit problem - February 3rd 2011, 01:02 AM

Okay. For your first question, let's use y = 1/x as an example. Even without graphing it, you should have a bit of an intuitive idea what it would look like. Now, try to find the right-hand limit as x --> 0. You'll find that you can't do it -- as x gets closer and closer to zero, y gets arbitrarily large. Now technically, a math major would tell you that the right-hand limit doesn't exist, and he'd be right. For the rest of the world though, who are not math majors, we'd just say that the right-hand limit is positive infinity. (It's important to remember that infinity isn't a number, that it isn't a "place" -- no matter how high you can go, you can always go higher.)

Now try to evaluate the left-hand limit. You run into a similar problem; the left-hand limit "does not exist" in a similar way, but we can just say that it's negative infinity. This becomes much clearer if you actually try graphing the function.

Notice that y is undefined for x = 0. x = 0 would be a vertical asymptote of the function.

So what is the limit as x --> 0? This limit doesn't exist, because it's positive infinity from one side and negative infinity for the other. This is what we call an infinite discontinuity, for pretty obvious reasons.

However, consider the function y = 1/x^2. In this case, both the right-hand and left-hand limits as x --> 0 are positive infinity, and so the limit as x --> 0 would be positive infinity, even though y(0) still doesn't exist and x = 0 is still an asymptote.

Depending on who you ask, vertical asymptotes may or may not be limits. I personally find it easier to consider the limits to be positive/negative infinity, but the limit being undefined is "more correct", because a limit is a boundary, and infinity really isn't a boundary.

As for your second question: for a continuous, everywhere differentiable function, yes -- it has to. However, if that function has a lot of jump or infinite discontinuities, then no.

Take y = int(x) as an example. y(4) can be clearly evaluated to equal 4, but the limit as x --> 4 does not exist.

As another example, take the piecewise function f(x) = {x^2 for x =/= 4, and 0 for x = 4}. The limit as x --> 4 would equal 16, but f(4) would be 0.

Any more questions? Ask away.


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Re: Calculus: Easy limit problem - February 3rd 2011, 01:10 AM

All right, so the next question would be "suppose the limit of a function as x appraoches 2 does NOT exist. Would that imply that the function is undefined at x=2?
   
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Re: Calculus: Easy limit problem - February 3rd 2011, 01:11 AM

No.

Try it, with y = int(x) as an example again.


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Re: Calculus: Easy limit problem - February 3rd 2011, 01:26 AM

Quote:
Originally Posted by Diaminohexanoate View Post
Say you have the function f(x) = (x - 3) ^2 / (x - 3). If you graph this function, it'll look exactly like the function g(x) = x - 3, with the exception of a removable discontinuity at (3,0)
Why would there be a removable discontinuity at 3,0?

This might be my last question.
   
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Re: Calculus: Easy limit problem - February 3rd 2011, 01:27 AM

Would the function be defined for x = 3?


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Re: Calculus: Easy limit problem - February 3rd 2011, 01:30 AM

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Originally Posted by Diaminohexanoate View Post
Would the function be defined for x = 3?
Well, it's asking that if the limit as x approaches 3 exists, would f(3) necessarily exist? So I'm assuming that x=3 is not defined? I'm not sure.
   
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Re: Calculus: Easy limit problem - February 3rd 2011, 01:31 AM

Well, let's make you sure.

f(x) = (x - 3) ^ 2 / (x - 3).

Try to evaluate f(3). What happens?


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Re: Calculus: Easy limit problem - February 3rd 2011, 01:35 AM

Quote:
Originally Posted by Diaminohexanoate View Post
Well, let's make you sure.

f(x) = (x - 3) ^ 2 / (x - 3).

Try to evaluate f(3). What happens?
The denominator would become undefined, so x=3 would not be defined in that case, so even if there's a removable discontinuity, there can still be a limit at y=0 when there's removable discontinuity?
   
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Re: Calculus: Easy limit problem - February 3rd 2011, 01:44 AM

Wait. When we're defining a function y in terms of x, we're not talking about the limit as y approaches something. x is the variable you're manipulating, not y. Just want to make sure you're clear about that.

Yes, the denominator would be zero which would make f(3) undefined. Good.

But look what happens when you take the right-hand limit. Suppose you take f(3.00000001). This would not be undefined; it would just be 00000001. You can make x arbitrarily close to 3, and f(x) will still be equal to x - 3. So as x approaches 3 from the right, it's clear that f(x) approaches 0.

Let's take the left-hand limit, then. Suppose you take f(2.99999999). This would still not be undefined; this would just be -0.000000001. You can again make x arbitrarily close to 3, and f(x) will still be equal to x - 3. So as x approaches 3 from the left, it's clear that f(x) approaches zero.

The fact that f(3) itself is undefined makes absolutely no difference to the behavior of f(x) around x = 3. Make sense?


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Re: Calculus: Easy limit problem - February 3rd 2011, 01:52 AM

Quote:
Originally Posted by Diaminohexanoate View Post
Wait. When we're defining a function y in terms of x, we're not talking about the limit as y approaches something. x is the variable you're manipulating, not y. Just want to make sure you're clear about that.

Yes, the denominator would be zero which would make f(3) undefined. Good.

But look what happens when you take the right-hand limit. Suppose you take f(3.00000001). This would not be undefined; it would just be 00000001. You can make x arbitrarily close to 3, and f(x) will still be equal to x - 3. So as x approaches 3 from the right, it's clear that f(x) approaches 0.

Let's take the left-hand limit, then. Suppose you take f(2.99999999). This would still not be undefined; this would just be -0.000000001. You can again make x arbitrarily close to 3, and f(x) will still be equal to x - 3. So as x approaches 3 from the left, it's clear that f(x) approaches zero.

The fact that f(3) itself is undefined makes absolutely no difference to the behavior of f(x) around x = 3. Make sense?
A little bit. I still don't understand everything completely, but I have the questions answered (even though I may not completely understand them lol) but this is a project grade and I'll learn more about limits as time progresses. I just gotta get used to it.

Let me just say that I really...really...really...appreciate your help. I was to the point that I was literally about to cry. I was very frustrated, and my sister wasn't doing a good job at explaining. But this definitely makes me feel better and I can pretend I know what I'm talking about for right now and then worry about that shit later.

Again, this means a lot to me. Thank you extremely much.
   
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Re: Calculus: Easy limit problem - February 3rd 2011, 02:00 AM

Hey,

No problem at all. I'm happy to have helped.

Honestly, limits are either intuitively obvious or very difficult to grasp. For me, limits were pretty intuitive, so I guess I lucked out a little for that one. I'll keep an eye out for something that can explain them better than I can.

Good luck on your projects and such! Feel free to ask away if you have any more questions.


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