$f(x)=\cos x-x$. If $\lim_{n\rightarrow \infty }|f(a_{n})|=0$, then $\lim_{n\rightarrow \infty }a_{n}$ converges?

Question

Here, we have a very well-known function $f(x)=\cos x-x$.

I know there is only one solution for $f(x)=0$.

Let it be $a\approx 0.739$.

Now, I'm trying to prove that if we have $a_{n+1}=\cos a_{n}$ for all $n$, then $a_{n}$ converges to $a$. (Just with MVT)

As a result, I got the fact "$\left | a_{n+2}-a_{n+1} \right |\leq \left | a_{n+1}-a_{n} \right |$ for all $n$."

So, I broke it into two cases s.t $\square $:"$\left | a_{n+2}-a_{n+1} \right |< \left | a_{n+1}-a_{n} \right |$ for any $n$," and $\bigcirc $:"$\left | a_{n+2}-a_{n+1} \right |= \left | a_{n+1}-a_{n} \right |$ for any $n$."

In case $\square $, we can get $\lim_{n\rightarrow \infty }|f(a_{n})|=0$.

But, I'm not certain if it is safe to say that "since $f(x)=0$ has a unique solution $x=a$, we have $\lim a_{n}=a$"

So, here's my general question: For any continuous & differentiable function $f(x)$ that has a unique solution $x=a$ for $f(x)=0$, if $\lim_{n\rightarrow \infty }|f(a_{n})|=0$, then is $\lim a_{n}=a$?

Answer 1

The answer is no. let $f(x)=xe^{x}$ Then $f(x)=0$ iff $x=0%$. But $f(-n) \to 0$ and $-n$ does not tend to $0$.