Problems of today's MATH202 tutorial notes #15

Problem 3. Let $\{a_n\}$ and $\{b_n\}$ be two sequences such that $\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=c$, suppose $f'(c)$ exists. Show that $\displaystyle \lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=f'(c)$.

My way.
I have to put some restictions to ensure my deduction is correct. If $f'(x)$ is continuous near $c$, then we are done with the Mean Value Theorem. But it, first and foremost, requires $f(x)$ to be differentiable.

To avoid continuity of derivative, we inevitably set Mean Value Theorem aside, then differentiability of $f(x)$ other than $c$ doesn't carry weight any more. But I have to suppose we always have $\boxed{a_n<c<b_n}$ or $\boxed{b_n<c<a_n}$ for sufficiently large $n$, this assumption will be used later on. By the fact that the existence of first order derivative implies the existance of linear approximation near $c$, we have for any $x$ near $c$, $$f(x)=f'(c)(x-c)+f(c)+o(x-c),$$ Here for any $\epsilon >0$, there exists a $\delta > 0$ such that $\displaystyle 0<|x-c|<\delta \implies \left|\frac{o(x-c)}{x-c}\right|<\frac{\epsilon}{2}$.

While for any $\delta_0 >0$, there exists an $N$ such that $n>N\implies |b_n-c|, |a_n-c|<\delta_0$, we take $\delta_0<\delta$, now for $n>N$, $$\begin{align*} &{\color{white}=}\left|\frac{f(b_n)-f(a_n)}{b_n-a_n}-f'(c)\right|\\ &=\left|\frac{f'(c)(b_n-c-(a_n-c))+o(b_n-c)+o(a_n-c)}{b_n-a_n}-f'(c)\right|\\ &=\left|\frac{o(b_n-c)+o(a_n-c)}{b_n-a_n}\right| \\ &\leq \left|\frac{o(b_n-c)}{b_n-c}\right|\left|\frac{b_n-c}{b_n-a_n}\right|+\left|\frac{o(a_n-c)}{a_n-c}\right|\left|\frac{a_n-c}{b_n-a_n}\right| \\ &<\left|\frac{o(b_n-c)}{b_n-c}\right|+\left|\frac{o(a_n-c)}{a_n-c}\right|\text{(by the assumption)}\\ &<\epsilon. \end{align*}$$

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Redo example 10. Show that $\displaystyle 2<65^{1/6}<2+\frac{1}{192}$.

My way.
The left hand side is obvious. For the right hand side, observe that $$65=2\left(1+\frac{1}{2^6}\right)^{\frac{1}{6}}<2\left(1+\frac{1}{6\cdot 2^6}\right)=2+\frac{1}{192}.$$ We have used $(1+x)^\alpha \leq 1+\alpha x,\forall x\ge 0,\forall \alpha \in[0,1]$.

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Problem 6. Let $f:[0,\infty)\to\mathbb{R}$ be continuous and $f(0)=0$. If $|f'(x)|<|f(x)|$ for very $x>0$, show that $f(x)=0$, for every $x\in[0,\infty)$.

My way.
Instead of considering $[0,\frac{1}{2}]$, we can consider any interval $[0,p]$, $p<1$ first. It can be shown that by mean value theorem, $$\begin{align*} |f(x)-f(0)|&<|f(y_n)||x|\\ &=|f(y_n)-f(0)||x|\\ &<|f(y_{n-1})||x||y_n|\\ &<\cdots \\ &<|x||y_n|\cdots |y_2||f(y_1)|\\ &<p^n|f(y_1)|,\forall x\in [0,p]. \end{align*}$$ Letting $n\to\infty$, we get $f(x)=\inf\{p^n|f(y_1)|:n\in\mathbb{N}\}=0$. We can show by induction that $f(x)=0$, for all $x\in [(n-1)p,np],n\in\mathbb{N}$ in the same manner.

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Problem 9. Give an example of a function that is differentiable on $(0,1)$ and $f(0)=f(1)$ but does not satisfy the Rolle's Theorem.

My way.
Define $f(x)=\sqrt{x},\forall x\in[0,1)$ and $f(1) = 0$.

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Additional problem. Let $f(x)$ be a differentiable on $[a,b]$, $f'(a)<f'(b),$ for any $y_0\in (f'(a),f'(b))$, prove that there exists $c\in(a,b)$ such that $f'(c)=y_0$.

My way.
Since that $f$ is differentiable cannot imply the continuity of $f'$. Intermediate value theorem fails to work here. We on the contrary define a new continuous function, $g(x) = f(x)-y_0 x$. It can be easily varified that $g'(a)=f'(a)-y_0<0$ and $g'(b)=f'(b)-y_0>0$, then extreme value cannot be attained at the end point, the minima must lie somewhere else in the interval $(a,b)$, hence there exists $c\in(a,b)$ such that $g'(c)=0\implies f'(c)=y_0$.