First L'Hospital Rule's proof (Bartle 4th ed)

Question

On the page 183, case (a) for example, how can we choose/construct such $c\in (a,b)$ and this $c$ is larger than or equal to $\beta.$

Any help would be appreciated.

Answer 1

Using the definition of $$ \lim_{x\to a^+} \frac {f'(x)}{g'(x)} = L, $$ i.e. $$ \forall \varepsilon > 0, \exists \delta > 0 \colon \left(a <x < a+\delta \implies L-\varepsilon < \frac {f'(x)}{g'(x)} < L+\varepsilon \right). $$ If this $\delta$ is large enough, your $c = a+\delta$ could be bigger than $\beta$. Although we do not require $c > \beta$.

UPDATE

$\alpha, \beta$ are arbitrary. In fact the proof could also be written as the following.

If $L \in \mathbb R$, then given $\varepsilon > 0$, there exists $c \in (a,b)$ s.t. $$u \in (a,c) \implies L-\varepsilon < \frac {f'(u)}{g'(u)} < L+\varepsilon. $$ Now for $a < \alpha < \beta \leqslant c$, by the Cauchy MVT we have $(2)$. Hence we have $(3)$.

Answer 2

You don't construct $c$ such that $c \geq \beta$. You find $c$ as in previous step. And then assert that if $\alpha$ and $\beta$ satisfy the inequalities $a<\alpha <\beta \leq c$ then $L-\epsilon <\frac {f(\beta)-f(\alpha)} {g(\beta)-g(\alpha)} <L+\epsilon$.