Proving that $xf'(x) = fracxx+1(x+1)f(x+1) - xf(x) - frac12xf''(ξ)$

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Let $f:(0,infty) to mathbb R$ be a function such that $f in C^2$, $lim_x to infty xf(x) = 2$ and $lim_x to infty xf''(x) = 0$. If $x>0$, prove that for some $ξ in (x,x+1)$ it is
$$xf'(x) = fracxx+1(x+1)f(x+1) - xf(x) - frac12xf''(ξ)$$
and then calculate the limit $lim_x to infty xf'(x)$.




Question/Request : I need some help, hints or tips on how to prove the formula in the center of the question. The final limit asked is simple as you just use the formula proven, so no help needed on that.






calculus differential-equations derivatives







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  • Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
    – Îµ-δ
    Sep 10 at 21:29










  • @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
    – Rebellos
    Sep 10 at 21:30










  • why do you write $fracxx+1(x+1)f(x+1)$ ?
    – Ahmad Bazzi
    Sep 10 at 21:33














up vote
0
down vote

favorite
2












Exercise :




Let $f:(0,infty) to mathbb R$ be a function such that $f in C^2$, $lim_x to infty xf(x) = 2$ and $lim_x to infty xf''(x) = 0$. If $x>0$, prove that for some $ξ in (x,x+1)$ it is
$$xf'(x) = fracxx+1(x+1)f(x+1) - xf(x) - frac12xf''(ξ)$$
and then calculate the limit $lim_x to infty xf'(x)$.




Question/Request : I need some help, hints or tips on how to prove the formula in the center of the question. The final limit asked is simple as you just use the formula proven, so no help needed on that.






calculus differential-equations derivatives







share|cite|improve this question





















  • Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
    – Îµ-δ
    Sep 10 at 21:29










  • @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
    – Rebellos
    Sep 10 at 21:30










  • why do you write $fracxx+1(x+1)f(x+1)$ ?
    – Ahmad Bazzi
    Sep 10 at 21:33












up vote
0
down vote

favorite
2









up vote
0
down vote

favorite
2






2





Exercise :




Let $f:(0,infty) to mathbb R$ be a function such that $f in C^2$, $lim_x to infty xf(x) = 2$ and $lim_x to infty xf''(x) = 0$. If $x>0$, prove that for some $ξ in (x,x+1)$ it is
$$xf'(x) = fracxx+1(x+1)f(x+1) - xf(x) - frac12xf''(ξ)$$
and then calculate the limit $lim_x to infty xf'(x)$.




Question/Request : I need some help, hints or tips on how to prove the formula in the center of the question. The final limit asked is simple as you just use the formula proven, so no help needed on that.






calculus differential-equations derivatives







share|cite|improve this question













Exercise :




Let $f:(0,infty) to mathbb R$ be a function such that $f in C^2$, $lim_x to infty xf(x) = 2$ and $lim_x to infty xf''(x) = 0$. If $x>0$, prove that for some $ξ in (x,x+1)$ it is
$$xf'(x) = fracxx+1(x+1)f(x+1) - xf(x) - frac12xf''(ξ)$$
and then calculate the limit $lim_x to infty xf'(x)$.




Question/Request : I need some help, hints or tips on how to prove the formula in the center of the question. The final limit asked is simple as you just use the formula proven, so no help needed on that.






calculus differential-equations derivatives




calculus differential-equations derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 10 at 21:25









Rebellos

10.2k21039




10.2k21039











  • Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
    – Îµ-δ
    Sep 10 at 21:29










  • @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
    – Rebellos
    Sep 10 at 21:30










  • why do you write $fracxx+1(x+1)f(x+1)$ ?
    – Ahmad Bazzi
    Sep 10 at 21:33
















  • Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
    – Îµ-δ
    Sep 10 at 21:29










  • @ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
    – Rebellos
    Sep 10 at 21:30










  • why do you write $fracxx+1(x+1)f(x+1)$ ?
    – Ahmad Bazzi
    Sep 10 at 21:33















Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
– Îµ-δ
Sep 10 at 21:29




Is the first term in the right-hand side what you meant to write? Because then $x+1$ would cancel out.
– Îµ-δ
Sep 10 at 21:29












@ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
– Rebellos
Sep 10 at 21:30




@ε-δ Was weird enough on when I was copying it from my textbook, but yes, according to it, this is the form of the expression.
– Rebellos
Sep 10 at 21:30












why do you write $fracxx+1(x+1)f(x+1)$ ?
– Ahmad Bazzi
Sep 10 at 21:33




why do you write $fracxx+1(x+1)f(x+1)$ ?
– Ahmad Bazzi
Sep 10 at 21:33










1 Answer
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I don't know why you would write the identity in this peculiar why, but this is just Taylor's theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f''(xi)$. Multiply by $x$ and you're done.






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    I don't know why you would write the identity in this peculiar why, but this is just Taylor's theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f''(xi)$. Multiply by $x$ and you're done.






    share|cite|improve this answer
























      up vote
      2
      down vote













      I don't know why you would write the identity in this peculiar why, but this is just Taylor's theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f''(xi)$. Multiply by $x$ and you're done.






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        I don't know why you would write the identity in this peculiar why, but this is just Taylor's theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f''(xi)$. Multiply by $x$ and you're done.






        share|cite|improve this answer












        I don't know why you would write the identity in this peculiar why, but this is just Taylor's theorem on the interval $[x,x+1]$. Since $fin C^2$, the theorem guarantees the existence of $xiin(x,x+1)$, so that $f(x+1)=f(x)+f'(x)+frac12f''(xi)$. Multiply by $x$ and you're done.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 10 at 22:12









        ε-δ

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