Proving a derivative exists given the limit of f' I have almost finished a problem from Baby Rudin, but I can't quite figure out the last step. The problem is as follows (5.9):
Let $f$ be:

*

*a continuous real function on $\mathbb{R}^{1}$,

*of which it is known that $f'(x)$ exists for all $x\neq0$, and

*$f'(x) \rightarrow 3$ as $x \rightarrow 0$.

Does it follow that $f'(0)$ exists?
Here's what I have so far:
By (1), there exists a $\delta_{1}$ with $|x|<\delta_{1} \rightarrow |f(x)-f(0)|<\frac{\epsilon}{3}$. By (2), there exists a $\delta_{2}$ with $|x-t|<\delta_{2}, x\neq0 \rightarrow |\frac{f(t)-f(x)}{t-x}-f'(x)|<\frac{\epsilon}{3}$. By (3), there exists a $\delta_{3}$ with $x\neq0, |x|<\delta_{3} \rightarrow |f'(x)-3|<\frac{\epsilon}{3}$.
So when $|x|<\min(\delta_{1},\delta_{3})$, $|x-t|<\min(\delta_{2},1)$:
$|\frac{f(t)-f(x)}{t-x}-f'(x)|+|f'(x)-3|+ |f(x)-f(0)| < \epsilon$
$|\frac{f(t)-f(x)}{t-x}-f'(x)+f'(x)-3+f(x)-f(0)| < \epsilon$
$|\frac{f(t)-f(x)}{t-x}-f'(x)+f'(x)-3+\frac{f(x)-f(0)}{t-x}| < \epsilon$
$|\frac{f(t)-f(0)}{t-x}-3|<\epsilon$
Which is very close to
$|\frac{f(t)-f(0)}{t}-3|<\epsilon$
which would yield a solution to the problem. But I can't figure out what condition to put on x or t-x to get from $|\frac{f(t)-f(0)}{t-x}-3|<\epsilon$ to $|\frac{f(t)-f(0)}{t}-3|<\epsilon$. Any help would be very, very appreciated.
 A: The cleanest approach (assuming you lack L'Hopital's rule) is to do as follows:
Given $\epsilon > 0$, take $\delta > 0$ such that for all $x$ where $0 < |x| < \delta$, $|f'(x) - 3| < \epsilon$. Then given any $t$ s.t. $0 < |t| < \delta$, we take by the MVT some $c$ between $t$ and $0$ such that $\frac{f(t) - f(0)}{t} = f'(c)$. Then $0 < |c| < \delta$, so $|f'(c) - 3| < \epsilon$, so $|\frac{f(t) - f(0)}{t} - 3| < \epsilon$.
This proves that $\lim\limits_{t \to 0} \frac{f(t) - f(0)}{t} = 3$. So $f'(0) = 3$.
This is basically the answer suggested by Snoop, but it does not rely on defining a function $y$ (which requires the axiom of choice).
A: The function is continuous on $\mathbb{R}$ and differentiable in any open interval that does not contain $0$. Thus by the MVT and taking the limit
$$\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}=\lim_{x \to 0}f'(y(x))$$
As $y(x) \to 0$ then $\lim_{y \to 0}f'(y)=3$. But by definition of the derivative
$$f'(0)=\lim_{x \to 0}\frac{f(x)-f(0)}{x-0}$$
as the limit exists and is $3$, then $f'(0)=3$.
A: You can use L'Hospital's rule:
$$
\lim_{t \to 0} \frac{f(t) - f(0)}{t} = \lim_{t \to 0} f'(t) , 
$$
and $\lim_{t \to 0} f'(t) = 3$ by assumption.  So we have
$$
\lim_{t \to 0} \frac{f(t) - f(0)}{t} = 3,
$$
and so $f'(0) = 3$.
A: If $h\geq 0$ then by Lagrange theorem
$$\left|\frac{f(h)-f(0)}{h} -3\right|\leq|f' (s) -3|\to 0$$ as $h\to 0.$
Analogously when $h\leq 0.$
