162 Proceedings of the Royal Society of Edinburgh. [Sess. 
Hence if one of them is a continuous function of the ensemble (x , y), so 
are they all , and they are equal. The same is true if one of them has a 
unique limit for all modes of approach to the point by means of points 
not on the axial cross through the point. 
PART III. 
§ 17. Lemma. — If f x is a continuous function of the ensemble (x , y) 
at the point (a , b), and f y exists at the point (a , b), then 
f (a + h , b + k) — f(a + li , b) 
IT 1 
has a unique double limit f y for all modes of approach of the point 
k 
(a + h, b + k) to the point (a, b) such that 
limits. 
For 
h 
has not zero for one of its 
f(a + h,h + k) r fia, + + b } ]( =/(a + /l)/ ,, h) _ f(a t h) 
h 1c 
_f(a + h , b + k) - f(a + h , b)j f{a + li. b) -f(a, b) ^ 
/l h 
Hence, using the Theorem of the Mean, 
f x (a + 6h,b + k)h + (f v (a, b) + y)/. + 1 U h ± k ) - f ( a + h ’ b h + (/,.(« , b) + v ’)h, 
lb 
0<6>< 1 , 
when 
0 being an otherwise unknown function of h and k ; is a function of k , 
but not of h , which vanishes with k ; is a function of h , but not of k , 
which vanishes with h . 
Therefore 
h 
k . 
Let us approach in such a way that 
f(a + h , b + k) -f(ci + h, b) 
k 
~f ,( a , h ) 
< ! f x (a + Oh , b + k) -ffa , b) 
h 
k 
+ Vk , + 
Vh 
k 
h 
( 1 ) 
where m is any positive quantity, finite or infinite. Then the right-hand 
side of the last inequality has the limit zero, therefore the same is true of 
the left-hand side. 
