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HW 13
Due Monday (Nov 29) 9pm. 
problem 6 contains a typo, and it is updated now.
1. In class we have seen that a function $f(x)$ may be differentiable everywhere, but the derivative function $f'(x)$ is not continuous. In this problem, we will see that discontinuity is not a removable singularity. Assume $f: \R \to \R$ is differentiable. Assume that $\lim_{x \to 0} f'(x) = 1$. Prove that $f'(0)=1$. (Hint, you can use mean value theorem, and definition of the $f'(0)$. )
2. If $f: [0,1] \to \R$ is a differentiable function such that $f'(0)=f'(1)=0$, is it true that there exists a $c \in (0,1)$, such that $f'( c) = 0$ as well? If not, give a counter-example.
3. Ross Ex 29.3
4. Ross Ex 29.5
5. Ross Ex 30.1
6. Prove that $\lim_{x \to 0^+} x^{-n} e^{-1/x} = 0$. Hint: let $u = 1/x$, and turn the problem into a $u \to \infty$ limit calculation, then Taylor expand $e^u = 1 + u + u^2/2! + \cdots + u^n / n! + \cdots $. (There was a typo in the first version, I wrote $x^{n} e^{-1/x}$ instead. You can either do the wrong problem, or do the corrected ones. )
Solution
HW 13
Due Monday (Nov 29) 9pm. problem 6 contains a typo, and it is updated now.
1. Assume that $\lim_{x \to 0} f'(x) = 1$. Prove that $f'(0)=1$. (Hint, you can use mean value theorem, and definition of the $f'(0)$. )
By mean value theorem for interval $[0,\delta]$, we have $$ \frac{f(\delta)-f(0)}{\delta - 0} = f'(x_\delta) $$ for some $x_\delta \in (0, \delta)$. Thus as $\delta \to 0$, $x_\delta \to 0$, hence $$ \lim_{\delta \to 0}\frac{f(\delta)-f(0)}{\delta - 0} = \lim_{\delta \to 0} f'(x_\delta) = \lim_{x \to 0} f'(x)1 $$
2. If $f: [0,1] \to \R$ is a differentiable function such that $f'(0)=f'(1)=0$, is it true that there exists a $c \in (0,1)$, such that $f'( c) = 0$ as well? If not, give a counter-example.
It may not be true. Let $g(x)= e^{-1/x}$ for $x \in (0,1]$ and $g(0)=0$. Then $g'(0)=0$. Let $f(x) = g(x) / (g(x) + g(1-x))$. We can prove that $f(x)$ is strictly monotone in $(0,1)$, and $f(0)=0, f(1)=1$.
3. Ross Ex 29.3
4. Ross Ex 29.5
5. Ross Ex 30.1
6. Prove that $\lim_{x \to 0^+} x^{-n} e^{-1/x} = 0$. Hint: let $u = 1/x$, and turn the problem into a $u \to \infty$ limit calculation, then Taylor expand $e^u = 1 + u + u^2/2! + \cdots + u^n / n! + \cdots $. (There was a typo in the first version, I wrote $x^{n} e^{-1/x}$ instead. You can either do the wrong problem, or do the corrected ones. )
