89] 
CONTINUOUS AND DISCONTINUOUS FUNCTIONS 
181 
number — m which — cf> (x) cannot surpass: or, what is the same 
thing, a greatest number in below which the value of cf) (x) cannot 
sink. In this case we say that cf)(x) is limited beloiu, and call m 
the lower limit of cf) (x). 
If both M and m can be determined in this way we shall say 
that cf) (x) is limited throughout (a, b). And then m and M are 
the greatest and least numbers respectively such that 
m i (f) (x) ^ M 
for all values of x in the interval (a, h). 
If M cannot be determined cf) (x) has no upper limit: in this 
case we can find values of cf) (x) algebraically greater than any 
number we can assign. In the same way <£ (x) may have no 
lower limit. 
Theorem 1. If cf)(x) is continuous throughout {a, h) it is 
limited throughout {a, h). 
We can certainly determine an interval (a, £), extending to 
the right from a, throughout which </>(#) is limited. For since 
<p{x) is continuous for x = a, we can, given any number B, however 
small, determine an interval (a, f) throughout which (f>{x) lies 
between cf) (a) — B and cf) (a) + B; and obviously cf) (x) is limited in 
this interval. 
Now divide the points £ of the interval (a, h) into two classes 
T. U, putting £ in T if <£(£) is limited in (a, £), and in U if this 
is not the case. By what precedes T certainly exists: what we 
propose to prove is that U does not. Suppose that U does exist; 
and let ¡3 be the value of £ which divides T from U. Since cf) (x) 
is continuous at x = ¡3 we can determine an interval (/3 — rj, /3 + v) 
throughout which cf> (¡3) — B < cf> (x) < cf) (¡3) + B, however small B 
may be. Thus cf) (x) is limited throughout (/3 — ij, /3 + y). But 
/3—7] belongs to T. Therefore <f>{x) is limited throughout (a, ¡3 — rj): 
and therefore it is limited throughout the whole interval (a, ¡3 + v). 
But /3 + 7] belongs to U and so cf> (x) is not limited throughout 
(a, /3 -f 7]). This contradiction shows that U does not exist. And 
so cf) (x) is limited throughout the whole interval (a, b). 
Theorem 2. If cf) (x) is continuous throughout (a, b) and M 
and m are its upper and lower limits, cf) (x) assumes the values M 
anfL m at least once each in the interval.