Limits : the answer is not obvious!

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  • #1
hermy
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i have been racking my brains on this question for a long time and am still unable to reach to the answer.

limit x → ∞ [(sin x + 2)/(sin x + 2) ] = ?

limit x → ∞ [ (sin x + 0.5)/(sin x + 0.5) ] = ?

the answers for both the questions are not the same.

there's also a hint: think in terms of imaginary no.s and argand plane
 
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  • #2
tiny-tim
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Hi hermy! :smile:

(have an infinity: ∞ :wink:)

I don't think it has anything to do with complex numbers …

it's more a question of whether the formula is defined for all values of x :wink:
 
  • #3
hermy
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it's more a question of whether the formula is defined for all values of x :wink:


since sin x can never be 2, the limit for sin x + 2........ must be 1.

but what puzzles me is why the limit shouldnt be 1 in the second case too?
 
  • #4
tiny-tim
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Because sinx can be -0.5 :wink:
 
  • #5
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Does it say that the limit for the second function is undefined?
 
  • #6
belliott4488
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I don't get this. For any value of x other than arcsin(-.5), isn't the function identically 1? If so, doesn't that mean that as you approach those singular values from either side, it remains 1, and does not diverge anywhere? A naive application of L'Hopital's rule would seem to suggest the same thing.

I would have said that both limits are identically 1, but I'm no mathematician.
 
  • #7
HallsofIvy
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I don't get this. For any value of x other than arcsin(-.5), isn't the function identically 1? If so, doesn't that mean that as you approach those singular values from either side,
What do you mean by "either side"? I've heard of "To infinity and beyond" but I was under the impression that there was not an "other side" to infinity!

it remains 1, and does not diverge anywhere? A naive application of L'Hopital's rule would seem to suggest the same thing.

I would have said that both limits are identically 1, but I'm no mathematician.
Were you thinking of the limit as x goes to 0?
 
  • #8
belliott4488
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What do you mean by "either side"? I've heard of "To infinity and beyond" but I was under the impression that there was not an "other side" to infinity!
Were you thinking of the limit as x goes to 0?
I meant as you approach any of the singular points, i.e. the points where sin(x) = -0.5. Everywhere else it's well-behaved, right? So if it's equal to 1 at these points and everywhere else, why does it change as you go to infinity?

In other words, I would have thought that both cases given would be equal in every way to
[tex]lim_{x\rightarrow\infty}[/tex]1

There must be some subtlety here that I'm missing.
 
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  • #9
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I was always of the impression that before you find limits, cancel anything that cancels, then rearange in order to get a form in which finding the limit is obvious... and in both of the limits you give it all cancels out. So either there is a typo, or you are asking for the limit as x tends to infinity of 1.

however in responce to the imaginary hint, sin (x) can be written as;

(eix-e-ix)/2i

where i = [tex]\sqrt{-1}[/tex]
 
  • #10
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I was always of the impression that before you find limits, cancel anything that cancels, then rearange in order to get a form in which finding the limit is obvious... and in both of the limits you give it all cancels out. So either there is a typo, or you are asking for the limit as x tends to infinity of 1.

however in responce to the imaginary hint, sin (x) can be written as;

(eix-e-ix)/2i

where i = [tex]\sqrt{-1}[/tex]
When you cancel, you are pulling out factors that are equal to 1. In the case of (sin x + .5)/(sin x + .5), that expression is equal to 1 as long as sin x is not equal to -.5.
 
  • #11
hermy
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The answer to the first question is 1 and the second limit is not defined. I have fair confidence in the authencity of the answer. (you could visit the website http://thepuzzlespot.blogspot.com/2008/08/perfect-in-limitshmmmm.html" [Broken] , where i found this question)

In my opinion, the explanation that sinx can be -0.5 does not suffice since x tends to infinity. sin x can be anything between -1 and 1. Isn't there any other way to go about the question?
 
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  • #12
tiny-tim
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Hi hermy! :smile:
The answer to the first question is 1 and the second limit is not defined.

That's correct:

the second expression is undefined again and again as you "approach infinity", so you can't find an N such that f(x) even exists for all x > N, let alone that it converges to something :wink:
 
  • #13
HallsofIvy
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In order to be able to say that "[itex]\lim_{x\rightarrow a} f(x)= L[/itex]", there must be some (punctured) neighborhood of a such that f(x) is defined for all x in that neighborhood. A "neighborhood of infinity" is a set {x> X} for some number X. No matter how large X is, there will be points in {x> X} for which (sin(x)- 0.5)/(sin(x)- 0.5) is not defined.
 
  • #14
belliott4488
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In order to be able to say that "[itex]\lim_{x\rightarrow a} f(x)= L[/itex]", there must be some (punctured) neighborhood of a such that f(x) is defined for all x in that neighborhood. A "neighborhood of infinity" is a set {x> X} for some number X. No matter how large X is, there will be points in {x> X} for which (sin(x)- 0.5)/(sin(x)- 0.5) is not defined.
I don't understand the statement "(sin(x)- 0.5)/(sin(x)- 0.5) is not defined".

This just looks to me like one of those first-year calculus problems where you are asked to determine the value of a ratio as both numerator and denominator go to zero. Is it zero? One? Infinite? We apply L'Hopital's rule to find the answer.

In this case, it seems unambiguous to me. First, however, may I restate the problem to refer instead to the ratio (sin(x)-1.0)/(sin(x)-1.0)? It's just easier to speak of the values x = n*pi/2 rather than arcsin(0.5).

EDIT: My mistake - please substitute "(4n+1)*pi/2" for "n*pi/2" throughout ... I was a little too hasty in my typing. (n is an integer, of course).

If I were asked what is the value of (sin(x)-1.0)/(sin(x)-1.0) for x -> n*pi/2, I would differentiate top and bottom, per L'Hopital, to get cox(x)/cos(x), and happily conclude that the answer is 1.

Is that incorrect? If not, and (sin(x)-1.0)/(sin(x)-1.0) is in fact equal to 1 for x = n*pi/2, then it is in fact equal to 1 for all x, in which case the limit as x goes to infinity is clear.

What's the subtlety that I'm missing? Does it have to do with the difference between asking the value of (sin(x)-1.0)/(sin(x)-1.0) as x approaches n*pi/2 versus the value for x = n*pi/2 identically? I admit that I don't generally distinguish between the two in practice.
 
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  • #15
jbunniii
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I don't understand the statement "(sin(x)- 0.5)/(sin(x)- 0.5) is not defined".

This just looks to me like one of those first-year calculus problems where you are asked to determine the value of a ratio as both numerator and denominator go to zero. Is it zero? One? Infinite? We apply L'Hopital's rule to find the answer.

The limit is being taken as x goes to INFINITY, not to arcsin(0.5). Neither the numerator nor the denominator has a limit as x goes to infinity, so L'Hopital's rule does not apply.

The statement "(sin(x)- 0.5)/(sin(x)- 0.5) is not defined" is true for any x that makes the denominator zero, and that happens with period 2*pi all the way out to infinity. That is why the limit does not exist.

Of course, if the domain were restricted to values of x that don't make the denominator zero, or if the function were explicitly defined to be 1 at those points, then the limit would exist, but neither appears to be the case.

That said, I have no idea what the original poster's hint ("think in terms of imaginary no.s and argand plane") has to do with this problem.
 
  • #16
belliott4488
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The limit is being taken as x goes to INFINITY, not to arcsin(0.5). Neither the numerator nor the denominator has a limit as x goes to infinity, so L'Hopital's rule does not apply.
I wasn't referring to that limit - I was referring to x = arcsin(0.5), which is where the function was asserted to be undefined. I understand that the limit as x goes to infinity is ill-defined if the function is undefined at these points, but I'm asking why it is undefined at these points.
The statement "(sin(x)- 0.5)/(sin(x)- 0.5) is not defined" is true for any x that makes the denominator zero, and that happens with period 2*pi all the way out to infinity. That is why the limit does not exist.
Of course a denominator of zero is problematic, but if the numerator is also zero, then you have to ask about the rate at which each one goes to zero as x approaches arcsin(0.5) - that's the basis for L'Hopital's rule, as I understand it. If the numerator goes to zero faster, then the ratio gets smaller as x approaches the limit, so we say the ratio goes to zero; if the denominator goes to zero faster, the ratio goes to infinity, and if they both go to zero at the same rate (as is the case here, since they're in fact equal), then the ratio goes to 1 - or in this case, it maintains a constant value of 1.

So my confusion persists: why do we have say that the ratio (sin(x)- 0.5)/(sin(x)- 0.5) is not defined for x = arcsin((0.5)? I would assert that the ratio is exactly equal to 1.0, for the reasons I just gave.
 
  • #17
tiny-tim
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So my confusion persists: why do we have say that the ratio (sin(x)- 0.5)/(sin(x)- 0.5) is not defined for x = arcsin((0.5)? I would assert that the ratio is exactly equal to 1.0, for the reasons I just gave.

Because when x = arcsin((0.5),

(sin(x)- 0.5)/(sin(x)- 0.5) = (0.5 - 0.5)/(0.5 - 0.5) = 0/0,

which isn't defined.
 
  • #18
jbunniii
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So my confusion persists: why do we have say that the ratio (sin(x)- 0.5)/(sin(x)- 0.5) is not defined for x = arcsin((0.5)? I would assert that the ratio is exactly equal to 1.0, for the reasons I just gave.

You have shown that the LIMIT of the ratio as x APPROACHES arcsin(0.5) is 1.0. That is not the same as saying that the ratio EQUALS 1.0 when x EQUALS arcsin(0.5). It doesn't.

If the definition was modified as follows:

[tex]f(x) = \left\{\begin{array}{ll}
1 & \textrm{if sin(x) = 0.5} \\
\frac{\sin(x) - 0.5}{\sin(x) - 0.5} & \textrm{otherwise} \end{array}[/tex]

Then it would be defined everywhere (in fact would equal 1 everywhere), so the limit would be 1.
 
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  • #19
belliott4488
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Because when x = arcsin((0.5),

(sin(x)- 0.5)/(sin(x)- 0.5) = (0.5 - 0.5)/(0.5 - 0.5) = 0/0,

which isn't defined.

Why would you not say the same thing about any other function to which you applied L'Hopital's Rule?

I thought that was the whole point - to disambiguate between the general rules:

0/anything = 0
anything/itself = 1
anything/0 = undefined
 
  • #20
belliott4488
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You have shown that the LIMIT of the ratio as x APPROACHES arcsin(0.5) is 1.0. That is not the same as saying that the ratio EQUALS 1.0 when x EQUALS arcsin(0.5). It doesn't.
I must be suffering from a very basic misconception, here.

Suppose I asked you what is the value of cos^2(x)/(sin(x)-1) at the point x = pi/2? Would you say this is undefined because the denominator is zero? I would not.

Two applications of L'Hopital's Rule show that this function does not diverge - in fact it goes to the value -2. You can easily plot it to see that. I've attached a plot of cos^2(x) (blue), sin(x)-1 (red), and cos^2(x)/(sin(x)-1) (magenta).

I need you guys to explain to me why L'Hopital's rule exists if you insist that a function must be undefined wherever its denominator goes to zero (regardless of the behavior of the numerator).
 

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  • #21
jbunniii
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I need you guys to explain to me why L'Hopital's rule exists if you insist that a function must be undefined wherever its denominator goes to zero (regardless of the behavior of the numerator).

L'Hopital's rule tells you what the LIMIT is, not what the function value is.

I could, if I liked, define a function

[tex]g(x) = \left\{\begin{array}{ll}
x/x & \textrm{if } x \neq 0 \\
5 & \textrm{if } x = 0 \end{array}[/tex]

L'Hopital's rule tells me that

[tex]\lim_{x \rightarrow 0} g(x) = 1[/tex]

but it does not follow that g(0) = 1.
 
  • #22
belliott4488
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Because when x = arcsin((0.5),

(sin(x)- 0.5)/(sin(x)- 0.5) = (0.5 - 0.5)/(0.5 - 0.5) = 0/0,

which isn't defined.
BTW, that statement is not strictly correct. 0/0 is not undefined, it's indeterminate.

Evaluating indeterminate forms is the purpose of L'Hopital's rule.

I assume you knew all that already, however, since it's basic Calculus and you're posting on this forum.
 
  • #23
belliott4488
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L'Hopital's rule tells you what the LIMIT is, not what the function value is.
The value of a function at a point where it becomes indeterminate is given by the limit. This is what Calculus is for - using limits to evaluate functions at points where they can't explicitly be evaluated. Your distinction baffles me.

I could, if I liked, define a function

[tex]g(x) = \left\{\begin{array}{ll}
x/x & \textrm{if } x \neq 0 \\
5 & \textrm{if } x = 0 \end{array}[/tex]
That's not a continuous function, and it does not have a well-defined derivative at x=5, so L'Hopital's Rules does not apply.

L'Hopital's rule tells me that

[tex]\lim_{x \rightarrow 0} g(x) = 1[/tex]
No, it doesn't, since the derivatives are not defined as you've defined the function. In fact, it's not even an indeterminate form, since you've avoided any occurrence of 0/0.
 
  • #24
jbunniii
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BTW, that statement is not strictly correct. 0/0 is not undefined, it's indeterminate.

Evaluating indeterminate forms is the purpose of L'Hopital's rule.

I assume you knew all that already, however, since it's basic Calculus and you're posting on this forum.

I think you should look up "indeterminate form". 0/0 is not defined.

An indeterminate form is something, for example, of the form

[tex]\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}[/tex]

where

[tex]\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} g(x) = 0[/tex] (these are what makes it an indeterminate form)

You dismiss this as "basic calculus" but you seem to overestimate your own mastery of the material.
 
  • #25
belliott4488
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I think you should look up "indeterminate form". 0/0 is not defined.

An indeterminate form is something, for example, of the form

[tex]\lim_{x \rightarrow 0}\frac{f(x)}{g(x)}[/tex]

where

[tex]\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} g(x) = 0[/tex] (these are what makes it an indeterminate form)

You dismiss this as "basic calculus" but you seem to overestimate your own mastery of the material.
I have looked up "indeterminate form", and 0/0 is given as a canonical example. See, for e.g. http://en.wikipedia.org/wiki/Indeterminate_form (where it's offered as a "common" example).

But I'm satisfied with the course of this discussion, now. I had thought I was missing something subtle, but I no longer think so.
 
  • #26
jbunniii
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That's not a continuous function, and it does not have a well-defined derivative at x=5, so L'Hopital's Rules does not apply.

I assume you mean "at x=0", but in any case the rest is bunk.

Have you read what L'Hopital's rule says? It does NOT require either continuity or differentiability at the limit point.

From, e.g., Apostol's "Calculus, Vol. 1":

Theorem 7.9. L'Hopital's rule for 0/0. Assume f and g have derivatives f'(x) and g'(x) at each point x of an open interval (a,b), and suppose that

[tex]\lim_{x\rightarrow a^+}f(x) = 0[/tex]

and

[tex]\lim_{x\rightarrow a^+}g(x) = 0[/tex]

Also assume that

[tex]g'(x) \neq 0[/tex]

for each x in (a,b). If the limit

[tex]\lim_{x \rightarrow a^+} \frac{f'(x)}{g'(x)}[/tex]

exists and has the value L, say, then the limit

[tex]\lim_{x \rightarrow a^+} \frac{f(x)}{g(x)}[/tex]

also exists and has the value L. [end statement of theorem]

Neither f'(a) nor g'(a) (nor for that matter, f(a) or g(a)) is required to exist.
 
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  • #27
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I must be suffering from a very basic misconception, here.
Yes. Another misconception you are suffering from is thinking that anything divided by itself equals 1. This is true except when the anything is zero. Also, the behavior of an expression is determined by whether the numerator is also zero or not. L'Hopital's rule is to be used for the indeterminate forms 0/0 and infinity/infinity, and provides a way to get limits of expressions with either of these two forms.
Suppose I asked you what is the value of cos^2(x)/(sin(x)-1) at the point x = pi/2? Would you say this is undefined because the denominator is zero? I would not.
This expression is undefined at x = pi/2 for precisely the reason you gave. Nevertheless, the limit of this expression exists as x approaches pi/2. As noted elsewhere in this thread, there's a difference between the value of an expression and its limit.
 
  • #28
belliott4488
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... As noted elsewhere in this thread, there's a difference between the value of an expression and its limit.
Okay, that's what I suggested earlier as the source of my trouble.

So it is wrong to say that the function I suggested earlier: cos^2(x)/(sin(x)-1) has a value of -2 for x = pi/2? It's pretty clear that it approaches that value from either side as you approach pi/2, so I've always thought it correct to say that the function itself takes that value at that point. Obviously we can't just plug x=pi/2 into the expression and evaluate it, but I had thought that the taking of the limit was an alternative way of determining the value of the function at that point. I gather this is incorrect?
 
  • #29
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The value of a function at a point where it becomes indeterminate is given by the limit.

This is absolutely not true. Where a function has a point discontinuity (a hole), L'Hopital's rule or other techniques can be used to determine the limit, and then you can define the function value to be equal to that limit in order to make the function continuous at that discontinuity.

As an example, f(x) = (x2 - 9)/(x - 3) is undefined at x = 3. Period. You can define f(3) = 6 to remove the discontinuity to make f define at x = 3, but until you do so, f is undefined at x = 3.
 
  • #30
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Okay, that's what I suggested earlier as the source of my trouble.

So it is wrong to say that the function I suggested earlier: cos^2(x)/(sin(x)-1) has a value of -2 for x = pi/2? It's pretty clear that it approaches that value from either side as you approach pi/2, so I've always thought it correct to say that the function itself takes that value at that point. Obviously we can't just plug x=pi/2 into the expression and evaluate it, but I had thought that the taking of the limit was an alternative way of determining the value of the function at that point. I gather this is incorrect?
Yes, it is incorrect. If the denominator of an expression is zero, the expression is undefined. It is immaterial whether the numerator happens to be zero or not. Either way, the expression is undefined.
 
  • #31
HallsofIvy
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All I can say is that Mark44 is completely correct. The limit is NOT necessarily the value of the function and, in particular, some thing like [itex](x^2- 4)/(x- 2)[/itex] is NOT defined at x= 2.
 
  • #32
belliott4488
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I assume you mean "at x=0", but in any case the rest is bunk.
Yes, I did mean that, sorry - but can you please tone down your language a bit? I'm just trying to understand this.

Have you read what L'Hopital's rule says? It does NOT require either continuity or differentiability at the limit point.
Yes, and the definition I read says that the limit of the ratio of the derivatives must exist as x approaches the limiting value, which I took to mean that the derivatives must exist at that point.
From, e.g., Apostol's "Calculus, Vol. 1":

Theorem 7.9. L'Hopital's rule for 0/0. Assume f and g have derivatives f'(x) and g'(x) at each point x of an open interval (a,b), and suppose that

[tex]\lim_{x\rightarrow a^+}f(x) = 0[/tex]

and

[tex]\lim_{x\rightarrow a^+}g(x) = 0[/tex]

Also assume that

[tex]g'(x) \neq 0[/tex]

for each x in (a,b). If the limit

[tex]\lim_{x \rightarrow a^+} \frac{f'(x)}{g'(x)}[/tex]

exists and has the value L, say, then the limit

[tex]\lim_{x \rightarrow a^+} \frac{f(x)}{g(x)}[/tex]

also exists and has the value L. [end statement of theorem]

Neither f'(a) nor g'(a) (nor for that matter, f(a) or g(a)) is required to exist.
Okay, so f'(x) and g'(x) must exist on the open interval (a,b), but not necessarily at x=a? I don't understand how L'Hopital works on a discontinuous function such as the one you've suggested. The proof I have in front of me uses the Cauchy Mean Value Theorem to prove L'Hopital, but does that apply to discontinuous functions?
 
  • #33
belliott4488
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All I can say is that Mark44 is completely correct. The limit is NOT necessarily the value of the function and, in particular, some thing like [itex](x^2- 4)/(x- 2)[/itex] is NOT defined at x= 2.
Okay, thanks, guys. As I said in my earlier posts, I suspected that I had a basic misunderstanding somewhere, and I guess this must be it.

I had thought that where a function is indeterminate, we simply can't say what its value is, but that we do not necessarily assert that it is discontinuous there. When you evaluate the limit at that point, I thought this was equivalent to evaluating the function at that point, but you're telling me that we must actually set the value of the function equal to the limit at that point, by hand, so to speak.

Okay, I think I've got it now.
 
  • #34
jbunniii
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Yes, I did mean that, sorry - but can you please tone down your language a bit? I'm just trying to understand this.

Sure, and I apologize for becoming a bit impatient. It sounded like you were telling us all that we didn't know what we were talking about and didn't understand basic calculus, but perhaps that was a misread on my part.

Yes, and the definition I read says that the limit of the ratio of the derivatives must exist as x approaches the limiting value, which I took to mean that the derivatives must exist at that point.

Okay, so f'(x) and g'(x) must exist on the open interval (a,b), but not necessarily at x=a? I don't understand how L'Hopital works on a discontinuous function such as the one you've suggested. The proof I have in front of me uses the Cauchy Mean Value Theorem to prove L'Hopital, but does that apply to discontinuous functions?

Right, you are free to change f(a) or g(a) or both, in any way you like, or even leave them completely unspecified, and the answer that L'Hopital gives will be the same.

I'm not sure of the specifics of your proof, but I bet it's stated in such a way that it never evaluates f or g at the point a, but only at points arbitrarily close to a. (i.e., on an interval of the form (a,b)). Under the hypotheses of L'Hopital's rule, the mean value theorem applies on (a,b), but not on any interval that INCLUDES the point a.
 
  • #35
Hurkyl
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In calculus, we usually don't actually care about the values of functions at individual points; it's the "bulk" behavior of the function that we're really interested in. So the thing we often do is to take the "continuous extension" of whatever partial function we have described. Unfortunately, this step doesn't often seem to be explicitly mentioned in introductory courses. :frown:


For example, consider the expression* f(x) := x/x. f(x) is constructed by taking the "diagonal" function [itex]\Delta(x) = (x, x)[/itex] with the "division" partial function [itex]q(x, y) = x/y[/itex].

q is only a partial function, because it's defined only for those (x, y) with y nonzero.

Alas, this means f is also merely a partial function, because the image of [itex]\Delta[/itex] does not lie in the domain of q. In particular, f is undefined at 0.

However, the limit of f at 0 exists; the "continuous" extension of f plugs this hole, and is the constant function 1.



*: x is a variable whose domain is all real numbers
 
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