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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.

  • 2
    First of all, thank you for posting an excellent model question on this site.$+1$. Second (and last) of all, I don't see what's wrong. – 2017-02-18
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    I think it must be a mistake in your book. E.g. taking $g(x)=x^2$, you get $f(x)=x$, and clearly $f'(0)\neq-1$ – 2017-02-18
  • 0
    Thank you both for your answers, I wasn't sure what to think anymore, i'm glad for your help. Edit: Thanks to the mod who fixed the post!. – 2017-02-18
  • 1
    You're correct; the book has a typo. – 2017-02-18

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