These  Real Analysis appendices continue the decimal viewpoint of limits, continuity and convergence in chapter 14. and this further lesson

A. What's Next
B. Pigeon Hole Principle
B. Bolzano-Weierstrass
C1. Triangle Inequality
C2. Triangle Inequality
C. More T.Inequality
D. Sets & Sequences
D. Monotone Sequences
E. Limits,  Properties
E Limits & Error Control
F. Continuous Functions
F. Closed Range Thm
F. Intermediate Val. Thm
F. Compactness Thm
F. Equicontinuity Thm
F Extreme Value Thm
G. Rolle's Theorem etc
G. Mean Val. Thm.
G. Constant Difference Thm
G. Lipschitz Continuity I
PS: One Sided Range Theorems
G. Velocity Revisited
G. Sufficient Conditions
H. Riemann Sums Conv
H. Lipschitz Continuity II

Proofs of  one-sided theorems could be of interest in the study of 2D topology.

Algebraic Properties of Limits

The above inequalities have the following consequences. 

Theorem E.2 [Algebraic Properties of Limits] Assume 

lim x->  a  f(x) = L    and    lim x->  a  g(x) = M.

where L and M are real numbers. Also assume k > 0 is a real number. Then 

lim x->  a  f(x)+g(x) = L+M lim x->  a  f(x)g(x) = LM lim x->  a  k·f(x) = kL

Moreover if M ¹ 0 then 

lim x->  a  f(x) g(x) = L M lim x->  a  1 g(x) = 1 M

Proof of the First Conclusion.

Observe 

|f(x)+g(x)-(L+M)| = |f(x)-L+(g(x)-M)| £ |f(x)-L|+|g(x)-M|

Suppose e > 0. There exists d₁ > 0 such that |x-a| < d₁ implies |f(x)-L| £ [1/2]e = e₁. Similarly, for the same e > 0, there exists d₂ > 0 such that |x-a| < d₂ implies |g(x)-M| £ [1/2]e = e₂. Now put d = min(d₁,d₂). Then |x-a| £ d implies both |f(x)-L| £ [1/2]e = e₁ and |g(x)-M| £ [1/2]e = e₂. The latter imply that

|f(x)+g(x)-L+M| £ |f(x)-L|+|g(x)-M|

£ e₁+e₂ = [1/2]e+[1/2]e = e. Since e > 0 is arbitrary, the foregoing implies

lim x->  a  f(x)+g(x) = L+M

Proof of the Second Conclusion.

Suppose e > 0. Recall from the inequality theorem that

|CD-cd| £ |C|e2+|D|e1+e1e2

if |c-C| £ e₁ and |d-D| £ e₂.

Therefore

|LM-f(x)g(x)| £ |L|e2+|M|e1+e1e2

if |f(x)-L| £ e₁ and |g(x)-M| £ e₂. Now if e₂ and e₁ are selected so that the first inequalities

1 3 e > |L|e2 1 3 e > |M|e1        and 1 3 e > e1e2

hold then

|LM-f(x)g(x)| £ e.

Now to satisfy the first inequalities, do the following.

1. Put e₂ = min(1,e[1/(|L|)]) if L ¹ 0 and put e₂ = 1 if L = 0. These choices imply the first inequality [1/3]e ³ e|L|e₂ and also 0 < e₂ £ 1
2. Put e₁ = [1/3]min(e,e[1/(|M|)]) if L ¹ 0 and put e₂ = [1/3] if M = 0. These choices imply the remaining two inequalities [1/3]e ³ |M|e₂ and [1/3]e ³ ee₂ since e₂ £ 1.

Now due to the hypotheses and the definition of limits, there exists d₁ > 0 such that |x-a| < d₁ implies |f(x)-L| £ e₁. Similarly, there exists d₂ > 0 such that |x-a| £ d₂ implies |g(x)-M| £ e₂. Now put d = min(d₁,d₂). Then |x-a| £ d implies both |f(x)-L| £ e₁ and |g(x)-M| £ e₂. The latter imply that |LM-f(x)g(x)| < e by the above reasoning. Since e > 0 is arbitrary, we are done.

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