12.5. LIMITS OF A FUNCTION 225

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|] . (12.8)

Now let 0 < δ 2 be such that if y ∈ D(f) and 0 < |x−y|< δ 2,

|f (y)−L|< ε

2(1+ |K|+ |L|), |g (y)−K|< ε

2(1+ |K|+ |L|).

Then letting 0 < δ ≤min(δ 1,δ 2), it follows from (12.8) that

|f ·g (y)−L ·K|< ε

and this proves (12.4).Consider (12.5). Let δ 1 be as above. From the properties of the cross product,

|f (y)×g (y)−L×K| ≤ |f (y)×g (y)−f (y)×K|+ |f (y)×K−L×K|

= |f (y)× (g (y)−K)|+ |(f (y)−L)×K|

Now from the geometric description of the cross product,

≤ |f (y)| |g (y)−K|+ |f (y)−L| |K|

Then if 0 < |y−x|< δ 1, this is no larger than

(1+ |L|) |g (y)−K|+ |f (y)−L| |K| ≤ (1+ |K|+ |L|) [|g (y)−K|+ |f (y)−L|]

and now the conclusion follows as before in the case of the dot product.The proof of (12.6) is left to you.Consider (12.7). Since h is continuous near L, it follows that for ε > 0 given, there

exists η > 0 such that if |y−L|< η , then

|h(y)−h (L)|< ε

Now since limy→xf (y) =L, there exists 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