Question about limit of function

[mathmari]

Hi! Knowing that the function \(\displaystyle g: (0,+\infty) \rightarrow R\) is differentiable and that \(\displaystyle lim_{x \rightarrow + \infty}{g'(x)}=0\), to show that \(\displaystyle lim_{x \rightarrow + \infty}{(g(x+1)-g(x))}=0\) could I use the mean value theorem in \(\displaystyle [x,x+1]\)?

 

[pka]

Hi! Knowing that the function \(\displaystyle g: (0,+\infty) \rightarrow R\) is differentiable and that \(\displaystyle lim_{x \rightarrow + \infty}{g'(x)}=0\), to show that \(\displaystyle lim_{x \rightarrow + \infty}{(g(x+1)-g(x))}=0\) could I use the mean value theorem in \(\displaystyle [x,x+1]\)?

If \(\displaystyle x > 0\) then \(\displaystyle \exists c_x\in(x,x+1)[g(x+1)-g(x)=g'(c_x)]~.\)

\(\displaystyle \displaystyle{\lim _{x \to \infty }}g'\left( {{c_x}} \right) = 0\) WHY?

 

[mathmari]

If \(\displaystyle x > 0\) then \(\displaystyle \exists c_x\in(x,x+1)[g(x+1)-g(x)=g'(c_x)]~.\)

\(\displaystyle \displaystyle{\lim _{x \to \infty }}g'\left( {{c_x}} \right) = 0\) WHY?

So can we not solve this exercise in this way?

 

[pka]

So can we not solve this exercise in this way?

Well, you tell us, why or why not.

 

[mathmari]

Well, you tell us, why or why not.

I thought that when we take the limit we could take \(\displaystyle c_x , x \rightarrow + \infty \).

 

[pka]

I thought that when we take the limit we could take \(\displaystyle c_x , x \rightarrow + \infty \).

We know that \(\displaystyle x < {c_x} < x + 1\) therefore \(\displaystyle \displaystyle{\lim _{x \to \infty }}{c_x} = \infty \). WHY?

If that is true then \(\displaystyle \displaystyle{\lim _{x \to \infty }}g'({c_x}) = 0\). Correct?

 

[mathmari]

We know that \(\displaystyle x < {c_x} < x + 1\) therefore \(\displaystyle \displaystyle{\lim _{x \to \infty }}{c_x} = \infty \). WHY?

If that is true then \(\displaystyle \displaystyle{\lim _{x \to \infty }}g'({c_x}) = 0\). Correct?

Since \(\displaystyle x < {c_x} < x + 1\) and \(\displaystyle \displaystyle{\lim _{x \to \infty }}{x} = \infty \) and \(\displaystyle \displaystyle{\lim _{x \to \infty }}{(x+1)} = \infty \) \(\displaystyle \Rightarrow \) \(\displaystyle \displaystyle{\lim _{x \to \infty }}{c_x} = \infty \).

 

[mathmari]

Mean value theorem states that

if f is continuous on a closed interval [a,b], b>a, and differentiable on the open interval (a,b) then there exists a point c in (a,b) such that

\(\displaystyle g'(c)=\displaystyle{\frac{g(b)-g(a)}{b-a}}\)

so as pka noted this implies that

\(\displaystyle \exists c \in (x,x+1) \ni g'(c)=\displaystyle{\frac{g(x+1)-g(x)}{(x+1)-x}}=g(x+1)-g(x)\)

see if you can finish from here. You had the right idea all along, you just need to follow through on it.

Can we continue as I said in my last two posts?

 

[pka]

Can we continue as I said in my last two posts?

Why are you still asking how to do this?
Reply #2 answered the question.

\(\displaystyle g(x+1)-g(x)=g'(c_x)\)

therefore \(\displaystyle \displaystyle{\lim _{x \to \infty }}\left[ {g(x + 1) - g(x)} \right] = {\lim _{x \to \infty }}g'({c_x}) = 0\)

 

[mathmari]

Ok! Thank you!!!!!!!!