10.8. LIMITS OF A FUNCTION 277

where K,L ∈W. Then if a, b ∈ F,

limy→x

(a f (y)+bg(y)) = aL+bK, (10.18)

If W is an inner product space,

limy→x

( f ,g)(y) = (L,K) (10.19)

If g is scalar valued with limy→x g(y) = K,

limy→x

f (y)g(y) = LK. (10.20)

Also, if h is a continuous function defined near L, then

limy→x

h◦ f (y) = h(L) . (10.21)

Suppose limy→x f (y) = L. If ∥ f (y)−b∥≤ r for all y sufficiently close to x, then |L−b| ≤ ralso.

Proof: Suppose 10.16. Then letting ε > 0 be given there exists δ > 0 such that if0 < ∥y− x∥< δ , it follows

| fk (y)−Lk| ≤ ∥f (y)−L∥< ε

which verifies 10.17.Now suppose 10.17 holds. Then letting ε > 0 be given, there exists δ k such that if

0 < ∥y− x∥< δ k, then| fk (y)−Lk|< ε.

Let 0 < δ < min(δ 1, · · · ,δ p). Then if 0 < ∥y− x∥< δ , it follows

∥f (y)−L∥∞< ε

Any other norm on Fm would work out the same way because the norms are all equivalent.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 forthese.

The proof of 10.18 is left for you. Now 10.19 is to be verified. Let ε > 0 be given.Then by the triangle inequality,

|( f ,g)(y)− (L,K)| ≤ |( f ,g)(y)− ( f (y) ,K)|+ |( f (y) ,K)− (L,K)|≤ ∥ f (y)∥∥g(y)−K∥+∥K∥∥ f (y)−L∥ .

There exists δ 1 such that if 0 < ∥y− x∥< δ 1 and y ∈ D( f ), then

∥ f (y)−L∥< 1,

and so for such y, the triangle inequality implies, ∥ f (y)∥ < 1+ ∥L∥. Therefore, for 0 <∥y− x∥< δ 1,

|( f ,g)(y)− (L,K)| ≤ (1+∥K∥+∥L∥) [∥g(y)−K∥+∥ f (y)−L∥] . (10.22)