So I am looking to see if my solution is correct. This is an exercise from Abbott's second edition of Understanding Analysis.
Exercise:
Let f be differentiable on an interval $A$ containing zero, and assume $(x_n)$ is a sequence in $A$ with $(x_n)→0$ and $x_n \neq 0$.
(a) If $f(x_n) = 0$ for all $n∈N$, show $f(0) = 0$ and $f'(0) = 0$.
(b) Add the assumption that f is twice-differentiable at zero and show that $f''(0) = 0$ as well.
My solution:
Here we let $(x_n) \rightarrow 0$ and $x_n \neq 0$
(a) Since $f$ is a continuous function on $A$. So $f(x_n) \rightarrow 0$ but $f(x_n) = 0$ for all $n \in N$. Therefore $f(0) = 0$.
Now $f'(0) = \lim_{x \rightarrow 0} ({f(x)- f(0)})/x = \lim_{x\rightarrow 0} f(x)/x$. As the function is differentiable on $A$, $f'(0)$ exists. Therefore $ \lim_{x \rightarrow 0} f(x)/x = \lim_{x_{n} \rightarrow \infty} f(x_{n})/x_{n} = \lim_{n \rightarrow \infty} 0/x_{n} = 0$. So we have $f'(0)=0$.
(b) If $f$ is twice differentiable then
$f''(0) = \lim_{x \rightarrow 0} ({f'(x)- f'(0)})/x = \lim_{x\rightarrow 0} f'(x)/x$. The limit exists by our assumption, so for any sequence $(y_n) \neq 0$ and $(y_n) \rightarrow 0$ we must have
$\lim_{n \rightarrow \infty} f'(y_n)/y_n = \lim_{n \rightarrow 0} f'(x)/x$. Now $\lim_{n \rightarrow \infty} f'(x_n)/x_n = \lim_{n \rightarrow \infty} 0/x_n = 0$. Therefore $f''(0)=0$.
Best Answer
It's all good except your last paragraph. You don't know that $f'(x_n) = 0$, so you can't rewrite $\lim_{n \rightarrow \infty} f'(x_n)/x_n$ as $\lim_{n \rightarrow \infty}0/x_n$. However, you can argue as follows:
Without loss of generality we can assume that for all $n$, $x_n\ne x_{n+1}$, and all $x_n$ are negative or all are positive. This is possible because either $\{n\in \mathbb{N} \mid x_n \gt 0\}$ or $\{n\in \mathbb{N} \mid x_n \gt 0\}$ is infinite. If $(x_n)$ doesn't have these properties, in any case it has a subsequence that does and we can use that.
For each $n$, $f(x_n) = f(x_{n+1}) = 0$, so by the mean value theorem there is $y_n$ between $x_n$ and $x_{n+1}$ such that $$ f'(y_n) = \frac {f(x_{n+1}) - f(x_n)} {x_{n+1} - x_n} = \frac 0 {x_{n+1} - x_n} = 0. $$ Clearly, $(y_n)\to 0$, and for all $n, y_n \ne 0$.
Because $f''$ exists at $0$, $$f''(0) = \lim_{n \rightarrow \infty} \frac {f'(y_n)}{y_n} = \lim_{n \rightarrow \infty} \frac 0{y_n} = 0. $$