Can Limits to Infinity Prove a Zero Derivative Over the Reals?

[JPBenowitz]

Suppose I have

lim_{t[itex]\rightarrow∞[/itex]} f(g(t)) = 0

and

lim_{t[itex]\rightarrow-∞[/itex]} f(g(t)) = 0

How would I prove [itex]\frac{df}{dt}[/itex][itex]_{|}\Re[/itex] = 0? (over the reals)

 

[theorem4.5.9]

This question doesn't make sense. Could you provide some context?

 

[JPBenowitz]

This question doesn't make sense. Could you provide some context?

I'm having problems displaying what I want to convey. Basically I proved that the limit for the curvature function will always converge to zero for any real continuous function and now I want to prove that the derivative with respect to time will always converge to zero.

So, essentially I need to prove that the integral over the reals of the curvature function will always converge to some constant.

 

[algebrat]

deleted

 

[blue_raver22]

To prove that \frac{df}{dt}_{|}\Re = 0, we can use the limit to derivative proof. First, we know that:

\lim_{t\rightarrow\infty} f(g(t)) = 0

This means that as t approaches infinity, the function f(g(t)) approaches 0. Similarly, we also know that:

\lim_{t\rightarrow-\infty} f(g(t)) = 0

This means that as t approaches negative infinity, the function f(g(t)) also approaches 0.

Now, we can use the definition of the derivative to find \frac{df}{dt}_{|}\Re:

\frac{df}{dt}_{|}\Re = \lim_{h\rightarrow0} \frac{f(g(t+h)) - f(g(t))}{h}

Since we are interested in the limit as t approaches infinity or negative infinity, we can substitute in \infty or -\infty for t in the above equation. This gives us:

\frac{df}{dt}_{|}\Re = \lim_{h\rightarrow0} \frac{f(g(\infty+h)) - f(g(\infty))}{h} = \lim_{h\rightarrow0} \frac{f(g(-\infty+h)) - f(g(-\infty))}{h}

Since we know that both f(g(\infty)) and f(g(-\infty)) are equal to 0, we can simplify the above equation to:

\frac{df}{dt}_{|}\Re = \lim_{h\rightarrow0} \frac{0 - 0}{h} = \lim_{h\rightarrow0} 0 = 0

Therefore, we have proven that \frac{df}{dt}_{|}\Re = 0, which means that the derivative of f with respect to t at any point on the real number line is equal to 0. This holds true for any real number, including infinity and negative infinity, which is what we were trying to prove.