Showing that $\lim f_n'=f'$ where $f_n(x)=\frac{nx^2+1}{2n+x}$ and $f = \lim f_n = \frac{x^2}{2}$

Question

I am self-learning Real Analysis from the text, Understanding Analysis by Stephen Abbott. I am having some troubles proving part c) of the below exercise problem. Any hints/suggestions without giving away the entire solution/proof would be extremely helpful.

[Abbott 6.3.5] Let

\begin{equation*} g_{n}( x) =\frac{nx+x^{2}}{2n} \end{equation*} and set $\displaystyle g( x) =\lim g_{n}( x)$. Show that $\displaystyle g$ is differentiable in two ways:

(a) Compute $\displaystyle g( x)$ by algebraically taking the limit as $\displaystyle n\rightarrow \infty $ and then find $\displaystyle g'( x)$.

Proof.

Fix $\displaystyle x\in \mathbf{R}$. We have: \begin{equation*} \begin{array}{ c l } \lim _{n\rightarrow \infty } g_{n}( x) & =\lim _{n\rightarrow \infty }\left(\frac{nx+x^{2}}{2n}\right)\\ & =\lim _{n\rightarrow \infty }\left(\frac{x+x^{2} /n}{2}\right)\\ & =\frac{\lim _{n\rightarrow \infty } x+\lim _{n\rightarrow \infty }\frac{x^{2}}{n}}{\lim _{n\rightarrow \infty } 2}\\ & =\frac{x}{2} \end{array} \end{equation*}

Thus, the limit function $\displaystyle g( x) =\frac{x}{2}$. The derivative of the limit function is:

\begin{equation*} g'( x) =\frac{1}{2} \end{equation*}

(b) Compute $\displaystyle g_{n} '( x)$ for each $\displaystyle n\in \mathbf{N}$ for each $\displaystyle n\in \mathbf{N}$ and show that the sequence of derivatives $\displaystyle ( g_{n} ')$ converges uniformly on every interval $\displaystyle [ -M,M]$. Use the theorem 6.3.3 to conclude that $\displaystyle g'( x) =\lim g_{n} '( x)$.

Proof.

Fix $\displaystyle n\in \mathbf{N}$. By the familiar rules of differentiation:

\begin{equation*} g_{n} '( x) =\frac{1}{2n}( n+2x) =\frac{1}{2} +\frac{x}{2n} \end{equation*} We are interested to prove that $\displaystyle g_{n} '( x)$ converges uniformly on any bounded interval $\displaystyle [ -M,M]$ to the constant function $\displaystyle h( x) =\frac{1}{2}$.

Pick an arbitrary $\displaystyle \epsilon >0$. We are interested to make the distance $\displaystyle |g_{n} '( x) -h( x) |$ smaller than $\displaystyle \epsilon $.

We have: \begin{equation*} \begin{array}{ c c } |g_{n} '( x) -h( x) | & =\left| \frac{x}{2n}\right| \\ & \leq \frac{M}{2n} \end{array} \end{equation*} If we pick $\displaystyle N >\frac{M}{2\epsilon }$, then for all $\displaystyle n\geq N$, and for all $\displaystyle x\in [ -M,M]$, $\displaystyle |g_{n} '( x) -g'( x) |< \epsilon $.

Consequently, the sequence of the derivatives $\displaystyle ( g_{n} ')$ converges uniformly on $\displaystyle [ -M,M]$ to the constant function $\displaystyle h( x) =\lim g_{n} '( x) =\frac{1}{2}$.

We find that the sequence of functions $\displaystyle ( g_{n})$ converge pointwise on the closed interval $\displaystyle [ -M,M]$ to $\displaystyle g$ and are differentiable. Since $\displaystyle ( g_{n} ')$ converges uniformly on $\displaystyle [ -M,M]$ to $\displaystyle h$, by the Differentiable Limit Theorem, it follows that $\displaystyle \lim g_{n} '=h=g'$ on $\displaystyle [ -M,M]$.

(c) Repeat parts (a) and (b) for the sequence $\displaystyle f_{n}( x) =\left( nx^{2} +1\right) /( 2n+x)$.

Proof.

Pointwise convergence of $\displaystyle f_{n}$:

Fix $\displaystyle x\in \mathbf{R}$. We have: \begin{equation*} \begin{array}{ c l } \lim _{n\rightarrow \infty } f_{n}( x) & =\lim _{n\rightarrow \infty }\frac{nx^{2} +1}{2n+x}\\ & =\lim _{n\rightarrow \infty }\frac{\left( x^{2} +1/n\right)}{( 2+x/n)}\\ & =\frac{x^{2}}{2} \end{array} \end{equation*}

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When I take the derivative $f_n'(x)$ and compute the expression $|f_n'(x)-f'(x)|$, I don't get something very tractable to find an upper bound. I would like to find an upper bound on $|f_n'(x)-f'(x)|$.

Answer 1

You have$$f_n'(x)-x=-\frac{3 n x^2+x^3+1}{(2 n+x)^2}.$$Take $M\in(0,\infty)$. Then, if $x\in[-M,M]$, you have$$|3nx^2+x^3+1|\leqslant3nM^2+M^3+1$$and, if $n>\frac M2$,$$(2n+x)^2\geqslant(2n-M)^2.$$Can you take it from here?

Answer 2

Uniform convergence of $\displaystyle f_{n} '$ on any bounded interval $\displaystyle [ -M,M]$:

Our claim is that the sequence of derivatives $\displaystyle ( f_{n} ')$ converges uniformly on $\displaystyle [ -M,M]$ to $\displaystyle g( x) =x$.

By the familiar rules of differentiation:

\begin{equation*} \begin{array}{ c l } f_{n} '( x) & =\frac{d}{dx}\left[\frac{nx^{2} +1}{2n+x}\right]\\ & =\frac{( 2n+x)( 2nx) -\left( nx^{2} +1\right)( 1)}{( 2n+x)^{2}}\\ & =\frac{nx^{2} +4n^{2} x-1}{( 2n+x)^{2}} \end{array} \end{equation*} We have:

\begin{equation*} \begin{array}{ c l } f_{n} '( x) -x & =\frac{nx^{2} +4n^{2} -1}{( 2n+x)^{2}} -x\\ & =\frac{nx^{2} +4n^{2} x-1-x\left( 4n^{2} +4nx+x^{2}\right)}{( 2n+x)^{2}}\\ & =\frac{nx^{2} +4n^{2} x-1-4n^{2} x-4nx^{2} -x^{3}}{( 2n+x)^{2}}\\ & =-\frac{x^{3} +3nx^{2} +1}{( 2n+x)^{2}} \end{array} \end{equation*} Pick an arbitrary $\displaystyle \epsilon >0$.

We have:

\begin{equation*} |f_{n} '( x) -x|=\frac{|x^{3} +3nx^{2} +1|}{( 2n+x)^{2}} \end{equation*} Since $\displaystyle x\in [ -M,M]$, we have: \begin{equation*} ( 2n+x)^{2} \geq ( 2n-M)^{2} \end{equation*} and

\begin{equation*} |x^{3} +3nx^{2} +1|\leq |x^{3} |+|3nx^{2} |+1\leq M^{3} +3nM^{2} +1 \end{equation*}

Pick $\displaystyle N_{1} >\max\left\{\frac{1}{M^{2}} ,M\right\}$. Then, for all $\displaystyle n\geq N_{1}$, since

\begin{equation*} n >M\Longrightarrow nM^{2} >M^{3} \ \Longrightarrow 3nM^{2} >M^{3} \quad \{M\in ( 0,\infty )\} \end{equation*} Also, \begin{equation*} nM^{2} >1 \end{equation*} Thus,

\begin{equation*} |x^{3} +3nx^{2} +1|\leq M^{3} +3nM^{2} +1\leq 3nM^{2} +3nM^{2} +nM^{2} =7nM^{2} \end{equation*}

Since, $\displaystyle M >0$, $\displaystyle 2n >2n-M$. Consequently:

\begin{equation*} \frac{1}{( 2n)^{2}} < \frac{1}{( 2n-M)^{2}} \end{equation*}

\begin{equation*} |f_{n} '( x) -x|=\frac{|x^{3} +3nx^{2} +1|}{( 2n+x)^{2}} \leq \frac{7nM^{2}}{4n^{2}} =\frac{7M^{2}}{4n} \end{equation*} Pick $\displaystyle N_{2} >\frac{7M^{2}}{4\epsilon }$. Then, for all $\displaystyle n\geq N_{2}$, the $\displaystyle |f_{n} '( x) -x|< \epsilon $.

Thus, a suitable response to the given $\displaystyle \epsilon $-challenge is $\displaystyle N=\max\{N_{1} ,N_{2}\}$.