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

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

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 $$|f_n'(x)-f'(x)|$$.

• Simply combining the two terms in the absolute value and noticing that the power of $n$ on the numerator and the denominator are respectively $1$ and $2$ may help. Oct 31, 2022 at 12:25

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?

• Using your suggestion, hope my work checks out. Oct 31, 2022 at 14:52

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 $$\displaystyle N=\max\{N_{1} ,N_{2}\}$$.