7.1. LIMITS OF A FUNCTION 185

Consider 7.6. Since h is continuous near L, it follows that for ε > 0 given, there existsη > 0 such that if ∥y−L∥< η , then ∥h(y)−h(L)∥< ε. Now since limy→x f (y) = L, thereexists δ > 0 such that if 0 < ∥y− x∥< δ , then ∥ f (y)−L∥< η .Therefore, if 0 < ∥y− x∥<δ , ∥h( f (y))−h(L)∥< ε.

It only remains to verify the last assertion. Assume ∥ f (y)−b∥ ≤ r. It is required toshow that ∥L−b∥ ≤ r. If this is not true, then ∥L−b∥ > r. Consider B(L,∥L−b∥− r).Since L is the limit of f , it follows f (y) ∈ B(L,∥L−b∥− r) whenever y ∈ D( f ) is closeenough to x. Thus, by the triangle inequality, ∥ f (y)−L∥< ∥L−b∥− r and so

r < ∥L−b∥−∥ f (y)−L∥ ≤ |∥b−L∥−∥ f (y)−L∥| ≤ ∥b− f (y)∥ ,

a contradiction to the assumption that ∥b− f (y)∥ ≤ r. ■The relation between continuity and limits is as follows.

Theorem 7.1.6 For f : D( f )→W and x ∈ D( f ) a limit point of D( f ), f is contin-uous at x if and only if limy→x f (y) = f (x) .

Proof: First suppose f is continuous at x a limit point of D( f ). Then for every ε > 0there exists δ > 0 such that if ∥x− y∥ < δ and y ∈ D( f ), then | f (x)− f (y)| < ε . Inparticular, this holds if 0 < ∥x− y∥ < δ and this is just the definition of the limit. Hencef (x) = limy→x f (y).

Next suppose x is a limit point of D( f ) and limy→x f (y) = f (x). This means that if ε >0 there exists δ > 0 such that for 0 < ∥x− y∥< δ and y ∈D( f ), it follows | f (y)− f (x)|<ε . However, if y = x, then | f (y)− f (x)| = | f (x)− f (x)| = 0 and so whenever y ∈ D( f )and ∥x− y∥< δ , it follows | f (x)− f (y)|< ε , showing f is continuous at x. ■

Example 7.1.7 Find lim(x,y)→(3,1)

(x2−9x−3 ,y

).

It is clear that lim(x,y)→(3,1)x2−9x−3 = 6 and lim(x,y)→(3,1) y= 1. Therefore, this limit equals

(6,1).

Example 7.1.8 Find lim(x,y)→(0,0)xy

x2+y2 .

First of all, observe the domain of the function is R2 \{(0,0)}, every point in R2 exceptthe origin. Therefore, (0,0) is a limit point of the domain of the function so it might makesense to take a limit. However, just as in the case of a function of one variable, the limit maynot exist. In fact, this is the case here. To see this, take points on the line y = 0. At thesepoints, the value of the function equals 0. Now consider points on the line y = x where thevalue of the function equals 1/2. Since, arbitrarily close to (0,0), there are points wherethe function equals 1/2 and points where the function has the value 0, it follows there canbe no limit. Just take ε = 1/10 for example. You cannot be within 1/10 of 1/2 and alsowithin 1/10 of 0 at the same time.

Note it is necessary to rely on the definition of the limit much more than in the case ofa function of one variable and there are no easy ways to do limit problems for functions ofmore than one variable. It is what it is and you will not deal with these concepts withoutsuffering and anguish.