Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is a proof in my book on introduction to real analysis I don't quite understand. I had to translate it so there could be some translation errors.

Theorem:

$f$ is differentiable in $a \iff$ $(\exists \alpha \in \mathbb{R})(\exists r:\mathbb{R} \rightarrow \mathbb{R}) : (f(a+h) = f(a) + \alpha h + hr(h) \;\text{with}\; r(h) \rightarrow 0 \;\text{if}\; h \rightarrow 0)$

Proof:

From the given follows: for $h \neq 0$ and sufficiently small, $\frac{f(a+h) - f(a)}{h} = \alpha + r(h)$. Conversely, suppose that $\lim_{h\rightarrow0}{\frac{f(a+h)-f(a)}{h}}$ exists and let $\alpha = \lim_{h\rightarrow0}{\frac{f(a+h)-f(a)}{h}}$. Then we can define, for $h \neq 0$ and sufficiently small: $r(h) = \frac{f(a+h)-f(a)}{h} - \alpha$. This means $f(a+h) = f(a) + \alpha h + hr(h) \;\text{with}\; r(h) \rightarrow 0 \;\text{if}\; h \rightarrow 0$

What I do not understand is the first sentence of the proof. How can that sentence prove that $\lim_{h\rightarrow0}{\frac{f(a+h)-f(a)}{h}}$ exists?

share|improve this question
add comment

1 Answer 1

up vote 1 down vote accepted

It exists because

$$ \lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h} = \lim_{h \rightarrow 0} \alpha + r(h) = \alpha. $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.