Doubt about a step in limits using mean value theorem

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Ozma

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Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] be a differentiable and bounded function. Show that if [imath]\lim_{x \to \infty}f'(x)[/imath] exists then [imath]\lim_{x \to \infty} f'(x)=0[/imath].

I tried this: let [imath]x>0[/imath], since [imath]f[/imath] is differentiable in [imath]\mathbb{R}[/imath] for the mean value theorem there exists [imath]c_x \in (x,2x)[/imath] such that [imath]f'(c_x)=\frac{f(2x)-f(x)}{x}[/imath]. But by hypothesis [imath]f[/imath] is bounded, so for the triangle inequality
[math]0 \le |f'(c_x)|=\left|\frac{f(2x)-f(x)}{x}\right| \le \frac{|f(2x)|+|f(x)|}{x} \le \frac{2M}{x}[/math]Letting [imath]x \to \infty[/imath], I deduce that [imath]\lim_{x \to \infty} |f'(c_x)|=0[/imath]. That is, [imath]\lim_{x \to \infty} f'(c_x)=0[/imath].

Question: I would like to deduce, from the facts that [imath]c_x \in (x,2x)[/imath], that [imath]\lim_{x \to \infty} f'(c_x)=0[/imath] and from the hypothesis [imath]\lim_{x \to \infty} f'(x)[/imath] exists that [imath]\lim_{x \to \infty} f'(x)=0[/imath]. For deducing this, can I let [imath]c_x=t[/imath] and notice that, when [imath]x \to \infty[/imath], from [imath]c_x \in (x,2x)[/imath] it follows that [imath]c_x \to \infty[/imath] as well and then [imath]0=\lim_{x \to \infty} f'(c_x)=\lim_{t \to \infty} f'(t) \implies \lim_{t \to \infty} f'(t)=0[/imath], and so conclude the proof?

Or similarly, can I argue that when the limit exists it is unique, so when [imath]x \to \infty[/imath] it is [imath]c_x \to \infty[/imath] and so the limits [imath]\lim_{x \to \infty} f'(c_x)[/imath] and [imath]\lim_{x \to \infty} f'(x)[/imath] must be equal by uniqueness since both [imath]x[/imath] and [imath]c_x[/imath] tends to [imath]\infty[/imath] as [imath]x \to \infty[/imath]?
 
Steven G said:
I'm still reading. You failed to define M.
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I am guessing that [imath]M[/imath] is the upper bound for [imath]|f(x)|[/imath], but I agree that it is a good idea to state this explicitly.
 
Ozma said:
Let [imath]f:\mathbb{R}\to\mathbb{R}[/imath] be a differentiable and bounded function. Show that if [imath]\lim_{x \to \infty}f'(x)[/imath] exists then [imath]\lim_{x \to \infty} f'(x)=0[/imath].

I tried this: let [imath]x>0[/imath], since [imath]f[/imath] is differentiable in [imath]\mathbb{R}[/imath] for the mean value theorem there exists [imath]c_x \in (x,2x)[/imath] such that [imath]f'(c_x)=\frac{f(2x)-f(x)}{x}[/imath]. But by hypothesis [imath]f[/imath] is bounded, so for the triangle inequality
[math]0 \le |f'(c_x)|=\left|\frac{f(2x)-f(x)}{x}\right| \le \frac{|f(2x)|+|f(x)|}{x} \le \frac{2M}{x}[/math]Letting [imath]x \to \infty[/imath], I deduce that [imath]\lim_{x \to \infty} |f'(c_x)|=0[/imath]. That is, [imath]\lim_{x \to \infty} f'(c_x)=0[/imath].

Question: I would like to deduce, from the facts that [imath]c_x \in (x,2x)[/imath], that [imath]\lim_{x \to \infty} f'(c_x)=0[/imath] and from the hypothesis [imath]\lim_{x \to \infty} f'(x)[/imath] exists that [imath]\lim_{x \to \infty} f'(x)=0[/imath]. For deducing this, can I let [imath]c_x=t[/imath] and notice that, when [imath]x \to \infty[/imath], from [imath]c_x \in (x,2x)[/imath] it follows that [imath]c_x \to \infty[/imath] as well and then [imath]0=\lim_{x \to \infty} f'(c_x)=\lim_{t \to \infty} f'(t) \implies \lim_{t \to \infty} f'(t)=0[/imath], and so conclude the proof?

Or similarly, can I argue that when the limit exists it is unique, so when [imath]x \to \infty[/imath] it is [imath]c_x \to \infty[/imath] and so the limits [imath]\lim_{x \to \infty} f'(c_x)[/imath] and [imath]\lim_{x \to \infty} f'(x)[/imath] must be equal by uniqueness since both [imath]x[/imath] and [imath]c_x[/imath] tends to [imath]\infty[/imath] as [imath]x \to \infty[/imath]?
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I believe you are on the right track, but this reasoning can be made cleaner and clearer: use proof by contradiction and assume that [imath]\lim_{x\rightarrow\infty} f^\prime(x) = a > 0[/imath]. Can you complete this approach?
 
Yes, [imath]M[/imath] is the constant that bounds [imath]f[/imath].

@blamocur: thanks for the suggestion, I will try that approach. However, my main trouble is to understand if my reasoning is correct because I don't want to go on in my studies with wrong beliefs. So it would be awesome if someone can check whether or not my proof works for what I asked after "Question", please :)
 
Ozma said:
Yes, [imath]M[/imath] is the constant that bounds [imath]f[/imath].

@blamocur: thanks for the suggestion, I will try that approach. However, my main trouble is to understand if my reasoning is correct because I don't want to go on in my studies with wrong beliefs. So it would be awesome if someone can check whether or not my proof works, please :)
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Honestly, I find your original post confusing. I could not find a formal proof there, and I cannot answer questions like "...can I argue...?". I suggest you write up a formal -- and preferably concise -- proof that we can look at and argue whether we consider the proof valid.
 
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