# Is this proof not sufficient?

Consider the following question

I would attempt to answer this as follows:

If the derivative of $f$ exists at $a \in (b,c)$, it would be defined as follows

$$f'(a) = \lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$$

And since $f(a) \equiv g(a)$ for all $a \in (b,c)$, we can say that

$$f'(a) \equiv \lim_{x \rightarrow a} \frac{g(x)-g(a)}{x-a} = g'(a) >$$

and since we are given that $g'(a)$ is defined for all $a \in (b,c)$, we know that $f'(a)$ must be defined for all such $a$.

However, the actual solution provided for this question is as below

So is my attempted proof insufficient? If so, why?

• You cannot say that $f'(x)=...$ as you don't know if it exists.To avoid this mistake start from $g'(x)$ and show that $\lim_{x \rightarrow a} \frac{f(x)-f(a)}{x-a}$ exists. Except for the above mentioned you are correct. – Andreas Ch. Feb 28 '17 at 0:43
• Many books on calculus try to over complicate things. The statement is too trivial to be proved explicitly and your thinking is correct in this matter. – Paramanand Singh Feb 28 '17 at 9:45

Now it begins to be so obvious that it is difficult to say what should be proven and what can be just stated. If we want to show that existence of limit is a local property, we expand the definition of limit and use $\delta < \min(a-b, c-a)$, so $f$ and $g$ restricted to $(a-\delta, a+\delta)$ are equal.