Properties of Limits - Proofs

NOTE:  You won't be responsible for this material in this class.   However, I still encourage all of you to read this over carefully.

Assume here that   Also assume k and r to be constants (fixed numbers).     Then the following theorems hold true:

Proof of [1]:   We'll prove the case with a "+", but the same steps are valid with a "–" instead.   

Suppose we’re given an e>0.   We must come up with a d that will guarantee that whenever  0<|x–a|< d,  it will be true that   | f(x)+g(x)–(L+M) | < e.

We are assuming that   Apply the definition of these limits, with epsilon equal to half our given e.  Let d be the smaller of the two deltas you get applying the definition to each of these limits.   Then, for   0<|x–a|< d,  we have:

QED

NOTE:   In the proof, we made use of this inequality:       

 

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Proof of [2]:  

Suppose we’re given an e>0.   We must come up with a d that will guarantee that whenever  0<|x–a|< d,  it will be true that | k f(x) – kL| < e.

We are assuming that   Apply the definition of this limit, with epsilon equal to our given e divided by |k|.  Let d be equal to the delta you get applying the definition to each of this limit.   Then, for   0<|x–a|< d,    we have:

QED

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Proof of [3]:  

Suppose we’re given an e>0.   We must come up with a d that will guarantee that whenever  0<|x–a|< d,  it will be true that  |f(x) g(x) – LM | < e.

We are assuming that   Apply the definition of these limits, with epsilon equal to half our given e.  Let d be the smaller of the two deltas you get applying the definition to each of these limits.   Then, for   0<|x–a|< d,  we have:

To see why this equality holds, just multiply out and simplify the expression in the absolute values on the right side of the equal sign.   Properties of absolute values allow us to deduce the inequalities:

At this point, we use the f  limit assumption with epsilon picked to be the minimum of

{ sqrt(e)/sqrt(3M), e/(3M) }.

Use the g limit assumption with epsilon picked to be the minimum of

{ sqrt(e)/sqrt(3), e/(3L) }. 

Now let epsilon be the smaller of these two epsilons.   Let  d  be the smaller of the deltas associated with this minimum epsilon.   Then our last inequality above is

QED