I was doing exercises from a book when I tried this particular one:
Find $f ^\prime (0)$ knowing that $g(0)=g^\prime (0)=0$ and $g^{\prime\prime}(0)=2$.
$$f(x)= \begin{cases} g(x)/x & \text{if $x$≠0} \\ 0 & \text{if $x$=0} \end{cases} $$
I used the definition of derivative:
$f^\prime(0)=\lim\limits_{x \to 0} \frac{f(x)-f(0)}{x-0}$
$f^\prime(0)=\lim\limits_{x \to 0} \frac{\frac{g(x)}x-0}{x-0}$
$f^\prime(0)=\lim\limits_{x \to 0} \frac{g(x)}{x^2}$
Since $g(0)=0$, I have and $0/0$ indetermination, So I applied L'Hopital:
$f ^\prime(0)=\lim\limits_{x \to 0} \frac{g^\prime(x)}{2x}$
Again, $0/0$ indetermination because $g^\prime(0)=0$, I applied L'Hopital one more time:
$f ^\prime(0)=\lim\limits_{x \to 0} \frac{g^{\prime\prime}(x)}{2}$
Now I got somewhere, $g^{\prime\prime}(0)=2$ so my limit equals $1$:
$f^\prime(0)=\lim\limits_{x \to 0} \frac{g^{\prime\prime}(0)}{2}=1$
However, answer says $f^\prime (0)= -1$ so I dont know what to think.
Did I miss something? Is this the correct way to solve this exercise?.
Thanks in advance for helping.