Proof that limit of difference of infinite limits is indeterminate

[swampwiz]

(NOTE: I have had a few similar postings lately on this subject, but they were much broader in scope, so I am posting only for this particular case; everything else has been figured out.)

If given that

lim_{x -> a} f( x ) = +∞

lim_{x -> a} g( x ) = +∞

what is the epsilon-delta formulation for the proof

lim_{x -> a} [ f( x ) - g( x ) ] is INDETERMINANT

?

 

[fresh_42]

No. It is not indeterminate, it is indeterminable.

 

[PeroK]

(NOTE: I have had a few similar postings lately on this subject, but they were much broader in scope, so I am posting only for this particular case; everything else has been figured out.)

If given that

lim_{x -> a} f( x ) = +∞

lim_{x -> a} g( x ) = +∞

what is the epsilon-delta formulation for the proof

lim_{x -> a} [ f( x ) - g( x ) ] is INDETERMINANT

?

What you need are examples of functions $f,g$ where the limit of the difference varies. You don't "prove" that sort of thing, per se, you find examples which differ in some key respect.

 

[Mark44]

To add to what @PeroK said, different pairs of functions can have very different limit values. For example, consider these three limits:
$f(x) = \frac 1 x$ and $g(x) = \frac 1 {x^2}$, and $\lim_{x \to 0^+}f(x) - g(x)$
$f(x) = \frac 1 {x^2}$ and $g(x) = \frac 1 x$, and $\lim_{x \to 0^+}f(x) - g(x)$
$f(x) = \frac 1 x$ and $g(x) = \frac 1 x$, and $\lim_{x \to 0^+}f(x) - g(x)$
To evaluate a limit using the definition of a limit (i.e., $\delta, \epsilon$), you have to have some idea in advance about whether the limit is finite or not.

 

[suremarc]

I don’t think a limit can be called indeterminate in and of itself. An “indeterminate form” according to Wikipedia is just a kind of expression obtained from a limit that behaves badly with respect to the algebraic limit theorem. In this case one does not say that $\lim_{x\rightarrow a}\,(f(x)-g(x))$ is indeterminate, but the expression obtained from $$\left(\lim_{x\rightarrow a}f(x)\right)-\left(\lim_{x\rightarrow a}g(x)\right)$$is an indeterminate form (the form being $\infty-\infty$). Depending on the choice of $f$ and $g$ one might get different results for $\lim_{x\rightarrow a}(f(x)-g(x))$, which means that we can’t uniquely define the (possibly infinite) value of an indeterminate form.

One could formalize this by showing any extended real number $L\in[-\infty,\infty]$ which we assign to $\infty-\infty$ fails to satisfy the following statement: $$(\forall f,g:\mathbb{R^R})(\forall a\in[-\infty,\infty])\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}g(x)=\infty\Rightarrow\lim_{x\rightarrow a}\left(f(x)-g(x)\right)=L.$$ This is a statement that essentially means we can consistently assign the value L to $\infty-\infty$, such that the algebraic limit theorem holds. To say that no value of L satisfies the statement means that for any value of L, there are functions $f,g$ which diverge to $\infty$ at some point $x=a$ such that $\lim_{x\rightarrow a}(f(x)-g(x))\neq L$. In other words, any attempt to define $\infty-\infty$ is inconsistent with the algebraic limit theorem.

 

[PAllen]

I am definitely more of a physics person than a math person, and I don't know if there is a name for the following approach, but the following makes sense to me as a proof (when formalized) that the type of limit proposed in the OP is indeterminate.

First consider f(x), g(x) where both have finite limits on approach to 0 (no loss of generality assuming zero), say L1 and L2, respectively. Then, it can be shown that for f(x)-g(x), for any epsilon around L1-L2, there is a delta such the every pair of x values within delta of zero produces f(x)-g(x) within epsilon of L1-L2.

In contrast, for any pair of functions f(x), g(x) such that each has infinite limit on approach to zero (including a case where f(x), g(x) are the same function e.g. 1/x), then the following is true:
for any delta around zero, there are pairs of points within that delta that produce any value value in (-∞,+∞) for f(x)-g(x).

That is, by allowing different points for evaluation of f and g, one sees a fundamental difference between a well defined limit and total indeterminacy.

 

[swampwiz]

I was able to finagle this expression into something that has a 0/0 term, which of course is the very essence of indeterminancy.

U = lim_{x -> a} f( x ) = +∞

V = lim_{x -> a} g( x ) = +∞

lim_{x -> a} ( [ e^{f( x )} ]^-1 ) = 0

lim_{x -> a} ( [ e^{g( x )} ]^-1 ) = 0

h( x ) = f( x ) - g( x )

lim_{x -> a} h( x ) = lim_{x -> a} [ ln ( e^{{ f( x ) - g( x ) }} ) ]

= lim_{x -> a} { ln ( [ e^{g( x )} ]^-1 / [ e^{f( x )} ]^-1 ) }

= ln { lim_{x -> a} ( [ e^{g( x )} ]^-1 / [ e^{f( x )} ]^-1 ) }

= ln ( { lim_{x -> a} [ e^{g( x )} ]^-1 } / { lim_{x -> a} [ e^{f( x )} ]^-1 } )

lim_{x -> a} h( x ) = - ln [ { 0 } / { 0 } ] -> INDETERMINATE FORM

 

[PeroK]

In contrast, for any pair of functions f(x), g(x) such that each has infinite limit on approach to zero (including a case where f(x), g(x) are the same function e.g. 1/x), then the following is true:
for any delta around zero, there are pairs of points within that delta that produce any value value in (-∞,+∞) for f(x)-g(x).

That is, by allowing different points for evaluation of f and g, one sees a fundamental difference between a well defined limit and total indeterminacy.

I can't say I'm convinced by this.

They key point is that the indeterminacy of the limit of $f(x)- g(x)$ in the case where $\lim_{x \rightarrow 0} f(x) = \lim_{x \rightarrow 0} g(x) = +\infty$ is shown by various examples. You can, of course, justify why counterexamples can be found; but, this is different from the techniques of a direct proof of something.

You might try to prove that
$$\lim_{x \rightarrow 0}(f(x) - g(x)) = 0$$
and fail to do this. But, you are not going to be able to show in general that
$$\lim_{x \rightarrow 0}(f(x) - g(x)) \ne 0$$
The limit might be $0$ and it might not be $0$.

More fundamentally, you might try to prove that the limit of $f(x) - g(x)$ exists (as an extended real number), but again you would find that you are unable to do this. The limit might exist and it might not. And, again, the proof of this is to find examples of both cases. If, for example:
$$f(x) = \frac{1}{|x|}, \ g(x) = \frac{1}{|x|} + \sin(\frac 1 x )$$
Then
$$\lim_{x \rightarrow 0} (f(x) - g(x)) \ \text{does not exist}$$
That example is the proof that the limit may not exist.

 

[WWGD]

If you mean that the difference between any two such functions is indeterminable then you have counterexamples like f(x)=2x and g(x)=3x whose limit exists and its infinity. Edit: for a more dramatic example consider a function and its Taylor series.

 

[Mark44]

If you mean <snip>

Presumably "you" means the OP, @swampwiz.

 

[zinq]

"If given that
limx -> a f( x ) = +∞
limx -> a g( x ) = +∞
what is the epsilon-delta formulation for the proof
limx -> a [ f( x ) - g( x ) ] is INDETERMINANT"

If that's all you know, you cannot determine the limit: It is indeterminate.

But it may be perfectly well determined if you knew what functions f and g are.

E.g. suppose that for all real numbers x ≠ 0, f and g are defined by f(x) = 1/x and g(x) = 1/x + 7. Then the limit of f(x) as x → 0 exists and, as is easy to see, is equal to -7.

 

[PeroK]

"If given that
limx -> a f( x ) = +∞
limx -> a g( x ) = +∞
what is the epsilon-delta formulation for the proof
limx -> a [ f( x ) - g( x ) ] is INDETERMINANT"

If that's all you know, you cannot determine the limit: It is indeterminate.

But it may be perfectly well determined if you knew what functions f and g are.

E.g. suppose that for all real numbers x ≠ 0, f and g are defined by f(x) = 1/x and g(x) = 1/x + 7. Then the limit of f(x) as x → 0 exists and, as is easy to see, is equal to -7.

Except that $\lim_{x \rightarrow 0} \frac 1 x \ne +\infty$, as the limit does not exist.

 

[zinq]

Sorry, PeroK, I'm not sure I understand. Are you saying that, because lim_x→0 1/x is not a real number, that the limit does not exist? Or rather that as x approaches 0, 1/x gets close to infinity only in absolute value? Maybe I should have used either lim_x→0+ 1/x, or else lim_x→0 1/x² ? I hope that takes care of it.

 

[PeroK]

Sorry, PeroK, I'm not sure I understand. Are you saying that, because lim_x→0 1/x is not a real number, that the limit does not exist? Fair enough. Should I have used either lim_x→0+ 1/x, or else lim_x→0 1/x² ? Okay, I should have used one of those.

$$\lim_{x \rightarrow 0^+} \frac 1 x = +\infty$$
$$\text{and} \ \lim_{x \rightarrow 0^-} \frac 1 x = -\infty $$

 

[zinq]

In other words, Yes.

 

[PeroK]

In other words, Yes.

In other words, the two-sided limit does not exist. And, it was two-sided limits that were under analysis.

 

[Mark44]

If given that
limx -> a f( x ) = +∞
limx -> a g( x ) = +∞

Your examples below don't agree with the given information. As already noted, $\lim_{x \to 0} f(x)$ and $\lim_{x \to 0} g(x)$ don't exist. The left- and right-side limits are as different as they could possibly be.

.g. suppose that for all real numbers x ≠ 0, f and g are defined by f(x) = 1/x and g(x) = 1/x + 7. Then the limit of f(x) as x → 0 exists and, as is easy to see, is equal to -7.

This example shows that it's possible for $\lim_{x \to 0}(f(x) - g(x)$ to exist even though the limits of the individual functions don't exist.

 

[swampwiz]

Just to be clear, I am not considering the case in which the 2 sides of the limit don't match (such as x^-1 @ 0); I am only considering a single-side limit for such a case.

 

[fresh_42]

Just to be clear, I am not considering the case in which the 2 sides of the limit don't match (such as x^-1 @ 0); I am only considering a single-side limit for such a case.

This doesn't matter. The answer to your questions has been given multiple times: You cannot draw conclusions, say, all imaginable outcomes have an example. There is no epsilontic in those cases, because epsilontic is only the analytic way to describe open neighborhoods. However, there is no open neighborhood around a point which does not exist! For infinite limits you must translate: "Increases beyond all barriers." and not invent a location at infinity. The question has been extensively answered, so it's time to close the thread.