AL Pure

It is because of condition (2), that states:
[h -> 0] lim f(h)/h = a, where a is real

Note:
f(x) is differentiable everywhere if f'(x) always exist; and …(i)
Left-hand limits and right-hand limits are the same iff ordinary limit exist. …(ii)

By condition (2), [h -> 0] lim f(h)/h = a where a is real, i.e. [h -> 0] lim f(h)/h exist.
That means, by (ii), [h -> 0+] lim f(h)/h = [h -> 0-] lim f(h)/h.

That means, once we found the ordinary limit exist, that is not necessary to prove the equality of left- and right-hand limit since they are already proved to be equal.

Since f'(x) = a[1 + f(x)] for all x and a are real numbers,
f'(x) is always exist, i.e. by (i), f(x) is differentiable everywhere.

lim{h->0+} (f(x+h)-f(x))/h and lim{h->0-} (f(x+h)-f(x))/h exist for any x
<=>
lim{h->0} (f(x+h)-f(x))/h exists for any x
<=>
f is diff. for any x


lim{h->0} .... has combined both left and right hand limits

as long as "two-hand" limit exists, both left and right land limits will exists too

Pak Wai is right

As f(x) is defined as a non-constant function and f:R->R, so f(x) is existed.
And so 1 + f(x) is also existed. As 'a' is not zero, a*[1 + f(x)] is also existed and is not equal to zero. f ' (x) = a*[1 + f(x)], so f ' (x) is existed and is not a constant. So f is differentiable everywhere. (Not an assumption.)

Hope you get what I mean.