Lim |f(x)| = +infty and f decreasing implies lim f(x) = +infty

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In summary: But, I think I understand now. Thanks again for the help!In summary, if f is defined on (a,b), \lim_{x \to a^{+}} |f(x)| = +\infty and f is decreasing on (a,b), then \lim_{x \to a^{+}} f(x) = +\infty as well. This is because for any chosen M > 0, there exists a \delta > 0 such that for all x \in (a,b) and a < x < a + \delta, f(x) > M or f(x) < -M. However, it can be shown that f(x) < -M cannot hold, which leads to a
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Homework Statement


If [tex]f[/tex] is defined on [tex](a,b)[/tex], [tex]\lim_{x \to a^{+}} |f(x)| = +\infty[/tex] and [tex]f[/tex] is decreasing on [tex](a,b)[/tex], show that [tex]\lim_{x \to a^{+}} f(x) = +\infty[/tex].

Homework Equations





The Attempt at a Solution


If [tex]\lim_{x \to a^{+}} |f(x)| = +\infty[/tex], then [tex]\forall M > 0, \exists \delta > 0[/tex] such that [tex]\forall x \in (a,b)[/tex] and [tex]a < x < a + \delta[/tex], we have [tex]|f(x)| > M[/tex], which implies we either have [tex]f(x) > M[/tex] or [tex]f(x) < -M[/tex]. It suffices to show that [tex]f(x) < -M[/tex] does not hold.

For contradiction, suppose that [tex]f(x) < -M[/tex] for [tex]x \in (a,b) \cap (a, a + \delta)[/tex]. Consider a sequence of real numbers [tex](x_n)[/tex] such that [tex]x_n \in (a,b) \cap (a, a + \delta)[/tex] and [tex]x_{n+1} > x_n[/tex] and [tex]\lim_{n \to \infty}x_n = a[/tex] for [tex]\forall n \in \mathbb{N}[/tex]. Since [tex]f[/tex] is decreasing, [tex]x_{n+1} > x_n[/tex] implies [tex]f(x_n) \geq f(x_{n+1})[/tex]. Thus, we have that [tex]-M > f(x_n) \geq f(x_{n+1})[/tex]...

But this is the part where I got stuck... any help appreciated! Thanks in advance!
 
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  • #2
It's always entirely possible that for some x really close to a, f(x) is really large in magnitude and negative. The key is that once you have that, f(x) can never get smaller than that. So for example, if you pick M cleverly, like

[tex]M=|f(\frac{a+b}{2})[/tex]|

then you know that for x<(a+b)/2 f(x) cannot be less than -M
 
  • #3
Office_Shredder said:
It's always entirely possible that for some x really close to a, f(x) is really large in magnitude and negative. The key is that once you have that, f(x) can never get smaller than that. So for example, if you pick M cleverly, like

[tex]M=|f(\frac{a+b}{2})[/tex]|

then you know that for x<(a+b)/2 f(x) cannot be less than -M

Thanks for the reply! I just want to follow up on this.

Suppose we have picked [tex]M=|f(\frac{a+b}{2})|[/tex], but since I want to show that [tex]f(x) < -M[/tex] cannot hold, I want to reach a contradiction somewhere. But I'm not seeing the contradiction. That is, suppose that we have [tex]f(x) < -M = -|f(\frac{a+b}{2})|[/tex] for [tex]x \in (a,b) \cap (a, a + \delta)[/tex]. But since [tex]f[/tex] is decreasing on [tex](a,b)[/tex], then we have for [tex]x < (a + b)/2, f(x) \geq f(\frac{a+b}{2})[/tex] (1)

But even with this, how can I show that [tex]f[/tex] cannot be less than [tex]-M[/tex]? That is, since [tex]M = |f(\frac{a+b}{2})|[/tex] is in terms of absolute values, but expression (1) does not exactly involve [tex]M[/tex] (i.e. no absolute values).

Any further help is appreciated! Thanks :)
 
  • #4
fmam3 said:
Thanks for the reply! I just want to follow up on this.

Suppose we have picked [tex]M=|f(\frac{a+b}{2})|[/tex], but since I want to show that [tex]f(x) < -M[/tex] cannot hold, I want to reach a contradiction somewhere. But I'm not seeing the contradiction. That is, suppose that we have [tex]f(x) < -M = -|f(\frac{a+b}{2})|[/tex] for [tex]x \in (a,b) \cap (a, a + \delta)[/tex]. But since [tex]f[/tex] is decreasing on [tex](a,b)[/tex], then we have for [tex]x < (a + b)/2, f(x) \geq f(\frac{a+b}{2})[/tex] (1)

But even with this, how can I show that [tex]f[/tex] cannot be less than [tex]-M[/tex]? That is, since [tex]M = |f(\frac{a+b}{2})|[/tex] is in terms of absolute values, but expression (1) does not exactly involve [tex]M[/tex] (i.e. no absolute values).

Any further help is appreciated! Thanks :)

[tex]f(\frac{a+b}{2})[/tex]

is either M or -M. I f(x)<-M, we have f(x)<M also (since M is positive). So regardless of whether [itex]f(\frac{a+b}{2}[/itex] is positive or negative, we have that f(x) is less than it.

But if [itex] x< \frac{a+b}{2}[/itex] it must be [itex] f(x) >f( \frac{a+b}{2})[/itex] as f is decreasing
 
  • #5
Office_Shredder said:
[tex]f(\frac{a+b}{2})[/tex]

is either M or -M. I f(x)<-M, we have f(x)<M also (since M is positive). So regardless of whether [itex]f(\frac{a+b}{2}[/itex] is positive or negative, we have that f(x) is less than it.

But if [itex] x< \frac{a+b}{2}[/itex] it must be [itex] f(x) >f( \frac{a+b}{2})[/itex] as f is decreasing

Thanks for the quick reply! I get the contradiction part in your last sentence, but I'm a little fuzzy on the first sentence. Just to be absolutely clear, I'll just spell it all out. Hope I'm understanding you correctly!

Let [tex]M > 0[/tex]. Then set [tex]M = |f(\frac{a+b}{2})|[/tex]. Then we must have that [tex]f(\frac{a+b}{2}) = M[/tex] (i.e. positive f) or [tex]f(\frac{a+b}{2}) = -M[/tex] (i.e. negative f). Now, we consider only the case [tex]f(x) < -M < 0[/tex]. Then, in this case, we must have the case of negative f. Hence, this would imply that [tex]f(x) < -M = -(-f(\frac{a+b}{2})) = f(\frac{a+b}{2})[/tex]. And from this, we use the fact that f is decreasing to reach the contradiction.

I just want to be sure that this is how I'm interpreting the below:
Office_Shredder said:
[tex]f(\frac{a+b}{2})[/tex] is either M or -M. I f(x)<-M, we have f(x)<M also (since M is positive). So regardless of whether [itex]f(\frac{a+b}{2}[/itex] is positive or negative, we have that f(x) is less than it.

Thanks again for the help!
 
  • #6
Actually, just to be nitpicking a bit. I think just saying [tex]x < (a + b) / 2[/tex] is not sufficient. Since the statement [tex]f(x) < -M[/tex] only holds for [tex]x \in (a,b) \cap (a, a + \delta)[/tex], it is not necessarily true that [tex](a + b) / 2[/tex] falls into that range. Hence, I think it would be better if we write, for [tex]a < x < \min\{(a+b)/2, a + \delta\}[/tex], then since f is decreasing on [tex](a,b)[/tex], it follows that [tex]f(x) \geq f(\frac{a+b}{2})[/tex] --- contradiction to the above.
 

1. What does "Lim |f(x)| = +infty" mean?

This notation means that the limit of the absolute value of the function f(x) approaches positive infinity as x approaches a certain value or as x approaches infinity.

2. What does it mean for a function to be decreasing?

A function is considered decreasing if its output values decrease as the input values increase. Visually, this would appear as a downward sloping line on a graph.

3. How does "Lim |f(x)| = +infty" relate to the function being decreasing?

If the limit of the absolute value of the function approaches positive infinity, it means that as x approaches a certain value or infinity, the function's output values are becoming increasingly large in a positive direction. This implies that the function is decreasing.

4. Can you give an example of a function that satisfies "Lim |f(x)| = +infty and f decreasing"?

One example of such a function is f(x) = -1/x. As x approaches 0 from the right, the limit of the absolute value of this function approaches positive infinity. Additionally, the function is decreasing as its output values decrease as x approaches 0 from the right.

5. Why is it important to understand "Lim |f(x)| = +infty and f decreasing implies lim f(x) = +infty"?

This concept is important because it helps us understand the behavior of a function as x approaches a certain value or infinity. It also allows us to make conclusions about the function's behavior based on its properties, such as being decreasing. This understanding is crucial in many areas of mathematics and science, particularly in analyzing and predicting the behavior of systems and processes.

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