0
$\begingroup$

This is from Khan Academy limits intro, it says that limit exists if $x$ approaches $7$ but I don't see how. When $x$ approaches $7$ we get different values from left and right side.

limits-graph

limits-answer

0 down vote accept

From the graph: g(8)=5,g(7)=2,g(5)=undefined,g(3)=3

if we come from right i.e.from g(8) then g(7) is approaching 4, g(x) is not defined for x=7 but it is not about having the value at x=7, it is about limits. If we approach x=7 from g(3) then g(x) approaches 2 and 2≠4. Where I am going wrong ?

$\endgroup$
  • 4
    $\begingroup$ What values do you see $g(x)$ approaching as $x$ approaches $7$ from each side? $\endgroup$ – G Tony Jacobs Aug 13 '18 at 3:18
  • 1
    $\begingroup$ The use of terminology is a bit odd. A limit is just a fixed value or it doesn't exist, it doesn't approach. Function values are the ones that do the approaching in some process $x\to a$. $\endgroup$ – Alvin Lepik Aug 13 '18 at 6:57
  • $\begingroup$ @AlvinLepik , I thought a limit means reasonable approximation. You are telling me it's a fixed value. Then another question arises, if limit is a fixed value then fixed value to 2 decimal places e.g. 0.25 or 4 decimal places 0.2563 or more ? $\endgroup$ – Arnuld Aug 13 '18 at 7:20
  • $\begingroup$ @Arnuld Right now I assume limits are unique if they exist (this isn't always true). If your limit is $\sqrt{2}$ then no approximation is sufficient. In your example, the numbers $0.2500$ and $0.2563$ are different, hence if the limit in some process is $0.2500$, it cannot be the other. $\endgroup$ – Alvin Lepik Aug 13 '18 at 8:18
  • $\begingroup$ @Arnuld A limit is indeed a 'fixed value', and is thought as the fixed value that one 'approaches' when they get this independent variable of approaching closer and closer to said limiting point. It doesn't 'change'. However the actual process of how one defines this 'fixed value' of a limit can be visualized as this whole 'approaching' business. $\endgroup$ – VgAcid Aug 13 '18 at 8:18
3
$\begingroup$

Are you sure about that? It looks like when you approach from the left and right, that it they approach the same 'holed' value. So, the limit as x goes to 7 approaches 4 and exists.

Note that the limit doesn't need to approach the actual value at that point either: that is the whole point of the concept of the 'limit', that which it 'approaches', not that which it is. Though, when the limit gives that which the variable 'approaches' is the 'right' value, it is called a continuous function.

$\endgroup$
  • $\begingroup$ you said limit doesn't need to approach the actual value at that point either: that is the whole point of the concept of the 'limit', that which it 'approaches', not that which it is My question is How do you notice this distinction ? How do you see it ? Sliding along the lines of the graph ? $\endgroup$ – Arnuld Aug 13 '18 at 5:11
  • $\begingroup$ One can see the limit intuitively by looking at a graph, and seeing how the 'line' of a function approaches a point, the point it approaches is the limit. How one notices this distinction is noting that just because it approaches a point, doesn't mean that the value at that point is correct or even exists. For example, the function f(x)=1/x, if I look at the 'line' of the function in the positive x, I will notice as x gets closer to 0(this can be thought of in the sliding business), 1/x goes to infinity. Yet, 1/0 is not defined. $\endgroup$ – VgAcid Aug 13 '18 at 8:23
  • $\begingroup$ As to how this distinction is truly made in mathematics, we can turn to the epsilon-delta definition. How exactly does one formulate this idea of 'approach'? Clearly, this concept of approach has to do with the fact that in a limit there is no smallest number between the difference of two 'y outputs'(the one we are limiting to and the one we are 'dragging' along), otherwise, there would be a clear break! So, for our output of our function f(x) and the value of the limit, we need for all e>0(that there is no smallest value of distance), |f(x)-L|(value of distance)<e. $\endgroup$ – VgAcid Aug 13 '18 at 8:44
  • $\begingroup$ Of course, w have 'glanced' over the whole approaching business on the x-axis. Clearly, the above can't occur for any fixed x, there has to be a condition on the x. We do however, want the above with specific e(for e=E, where E is a number, |f(x)-L|<E) to occur when x is only the right amount of close to c but not at c, that is, 0<|x-c|<d. Or, putting it together, for all e>0, that there exists d>0 such that 0<|x-c|<d means |f(x)-L|<e. c is the limiting point, and L our limit of f(x) as x goes to c, and we use this to formally notice the distinction. $\endgroup$ – VgAcid Aug 13 '18 at 8:55
1
$\begingroup$

From the left, as $x \to 7^-$, it gets closer and closer to $4$.

From the right, as $x \to 7^+$, it gets closer and closer to $4$ as well.

Hence the limit is $4$.

Edit:

What matters is value that is very close around $7$, check out the value like $g(6.9)$ , $g(6.99),g(6.999)$ and so on, they are close to $4$, not $2$ right?

$\endgroup$
  • $\begingroup$ I have edited the OP for clairty $\endgroup$ – Arnuld Aug 13 '18 at 5:09
  • $\begingroup$ edited: i don't see how you get the value $2$. $\endgroup$ – Siong Thye Goh Aug 13 '18 at 5:14
0
$\begingroup$

I think I found the answer after a little hint by VgAcid above. Notice 2 things, 1st is explanation by VgAcid and 2nd is noticing graphs of chapter on limits :

1) the limit of g at x=3, equals, 3 is equal to 5, but the value of g at x=3 is undefined! They are not the same! That's the beauty of limits: they don't depend on the actual value of the function at the limit. They describe how the function behaves when it gets close to the limit. It means closed-circle (value not defined) does not affect the limits in anyway.

2) We find limits by sliding-along the graph or walking-on the lines of the graph, never by matching x values to y (this is what I was doing). When you slide-along or walk-on the lines of the graph you see from x = 9 to x = 8 and then towards x = 7, graph-line is moving towards y = 4 , graph-line moving towards y = 4 means x is approaching y = 4 (it does not matter if that y = 4 value is excluded, you have to see where the lines on the graph go)

P.S. I think the culprit (at least the associate of culprit) for this confusion is the equal sign (symbol ?) $=$. When we are using $x$ with $limits$ we are using $x \to 7$ but when we are using $g(x)$ we are using $=$: $$ x \to 7, g(x) = $$

We read it as as x approaches 7 then g(x) is equal to , hence the confusion. g(x) may or may not be equal to the value. 2nd, we are interested in $limits$ (opposed to values). $x$ and $g(x)$ are doing the same thing, approaching some value, but using different notation for same cause is the reason behind this confusion. It would have been much better if it had been something like this:

$$ x \to 7, g(x) \to $$

But who I am to say this. Mathematical notation, whatever it is, exists for a reason. Mathematics always a synonym for clarity and terseness. There must be a reason behind using $=$, in same expression, for both assigning values and approaching values.

$\endgroup$
  • 1
    $\begingroup$ Please edit your original question to clarify it - only use the answer box to actually answer the question. $\endgroup$ – user296602 Aug 13 '18 at 5:07
  • $\begingroup$ Close, I think you are getting the idea better. The left limit at x=3(that is, what the 'left line' approaches in the y value as x goes to 3 from the left) is 3. The right limit at x=3 is 6(I assume you meant 6, not 5). However, the value at g(3) is completely defined. It is, per the graph, the solid dot, to be 3, the open dot signifies that there ISN'T a point there. Maybe what you mean to say is that the value of the 'both sided limits' is undefined, since the left approaches 3 and the right approaches 5? $\endgroup$ – VgAcid Aug 13 '18 at 8:27
  • $\begingroup$ Everything else in terms of understanding looks correct. In terms of notation, I can kind of see where you are getting at, though it isn't 100% correct. Remember, the actual thing the function isn't 'affected' by our conclusion of limits. As x approaches 7, g(x) is unaffected. The value of the limit, or lim g(x) is affected however. This 'lim' notation considers the concept of 'approaching', the g(x)= represents the actual value. Your notation of x -> 7, g(x) -> I don't see any obvious problems with. $\endgroup$ – VgAcid Aug 13 '18 at 8:36
  • $\begingroup$ Actually, on the notation, we could write in general, using your notation, x->a, g(x)-> b, which would contain the same amount of possible information as [lim x-> a g(x)] = b. The reason we use the former is equality is more familiar, and, makes a more clear distinction of the input variable being limited and the output of what it is going to. $\endgroup$ – VgAcid Aug 13 '18 at 8:40
  • $\begingroup$ I got your point. Limit is a fixed value, so of course, equality symbol is used. Thank you so much $\endgroup$ – Arnuld Aug 13 '18 at 8:43
0
$\begingroup$

Algebraically, note that: $$g(x)=\begin{cases}x-3, x\in[6,7)\cup (7,9]\\ \ \ \ \ \ \ \ 2, x=7\end{cases}$$ Now take the limit: $$\lim_{x\to 7^-} g(x)=7-3=4;\\ \lim_{x\to 7^+} g(x)=7-3=4;\\ \lim_{x\to 7} g(x)=\lim_{x\to 7^-} g(x)=\lim_{x\to 7^+} g(x)=4.$$

$\delta-\epsilon$ notation: $$\lim_{x\to a} g(x)=L \iff \\ \forall \epsilon>0, \exists \delta>0 \ \ \text{such that} \ \ 0<|x-a|<\delta \Rightarrow |g(x)-L|<\epsilon$$

It is: $$\lim_{x\to 7} g(x)=4 \iff \\ \forall \epsilon>0, \exists \delta>0 \ \ \text{such that} \ \ 0<|x-7|<\delta \Rightarrow |x-3-4|=|x-7|<\delta=\epsilon.$$ Note that: $$\lim_{x\to 7} g(x)=4\ne g(7)=2$$ therefore $g(x)$ is not continuous at $x=7$.

See more on Limit of a function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.