As in the case of scalar valued functions of one variable, a concept closely related to continuity is that of the limit of a function. The notion of limit of a function makes sense at points x, which are limit points of D
Definition 15.8.1 Let A ⊆ ℝ^{m} be a set. A point x, is a limit point of A if B
Definition 15.8.2 Let f : D

if and only if the following condition holds. For all ε > 0 there exists δ > 0 such that if

then,

Theorem 15.8.3 If lim_{y→x}f
Proof: Let ε > 0 be given. There exists δ > 0 such that if 0 <

Pick such a y. There exists one because x is a limit point of D

Since ε > 0 was arbitrary, this shows L = L_{1}. ■
As in the case of functions of one variable, one can define what it means for lim_{y→x}f
Definition 15.8.4 If f
The following theorem is just like the one variable version of calculus.
 (15.2) 
if and only if
 (15.3) 
where f

where K,L ∈ ℝ^{q}. Then if a, b ∈ ℝ,
 (15.4) 
 (15.5) 
In the case where q = 3 and lim_{y→x}f
 (15.6) 
If g is scalar valued with lim_{y→x}g
 (15.7) 
Also, if h is a continuous function defined near L, then
 (15.8) 
Suppose lim_{y→x}f
Proof: Suppose 15.2. Then letting ε > 0 be given there exists δ > 0 such that if 0 <

which verifies 15.3.
Now suppose 15.3 holds. Then letting ε > 0 be given, there exists δ_{k} such that if 0 <

Let 0 < δ < min

Each of the remaining assertions follows immediately from the coordinate descriptions of the various expressions and the first part. However, I will give a different argument for these.
The proof of 15.4 is left for you. It is like a corresponding theorem for continuous functions. Now 15.5 is to be verified. Let ε > 0 be given. Then by the triangle inequality followed by the Cauchy Schwarz inequality,
≤ 

+ 

≤ +
. 
There exists δ_{1} such that if 0 <

and so for such y, the triangle inequality implies,
 (15.9) 
Now let 0 < δ_{2} be such that if y ∈ D

Then letting 0 < δ ≤ min

and this proves 15.5.
Consider 15.6. Let δ_{1} be as above. From the properties of the cross product,


Now from the geometric description of the cross product,

Then if 0 <

and now the conclusion follows as before in the case of the dot product.
The proof of 15.7 is left to you.
Consider 15.8. Since h is continuous near L, it follows that for ε > 0 given, there exists η > 0 such that if

Now since lim_{y→x}f

Therefore, if 0 <

It only remains to verify the last assertion. Assume

and so

a contradiction to the assumption that
The relation between continuity and limits is as follows.
Proof: First suppose f is continuous at x a limit point of D
Next suppose x is a limit point of D
Example 15.8.7 Find lim_{}
It is clear that lim_{}
Example 15.8.8 Find lim_{}
First of all, observe the domain of the function is ℝ^{2} ∖
Note it is necessary to rely on the definition of the limit much more than in the case of a function of one variable and there are no easy ways to do limit problems for functions of more than one variable. It is what it is and you will not deal with these concepts without suffering and anguish.