# confused about delta epsilon proofs

• Jan 1st 2010, 03:05 PM
oblixps
i know how to do basic proofs, but some proofs on the actual limit theorems confuse me. my textbook's choices for delta are very obscure and i have no idea how they even came up with that.

for the proof of the limit theorem where the limit of a product of 2 functions is equal to the product of their limits, my book did: f = L1 + (f-L1) and g = L2 + (g-L2). and they want to show that |fg - L1*L2| < ε if 0<|x-a|<δ.

so with substitution and rearrangement they get |L1(g-L2)+L2(f-L1)+(f-L1)(g-L2)|< ε. since the limit of f as x approaches a is L1 and limit of g as x approaches a is L2, we can find positive numbers δ1, δ2, δ3, δ4 such that:

|f-L1|< sqrt(ε/3) if 0<|x-a|<δ1
|f-L1|< ε/[3(1+|L2|)] if 0<|x-a|<δ2
|g-L2|< sqrt(ε/3) if 0<|x-a|<δ3
|g-L2|< ε/[3(1+|L1|)] if 0<|x-a|<δ4

the rest of the proof i understand but what confuses me is where and how did they get those expressions sqrt(ε/3) and ε/[3(1+|L2|)]?
• Jan 4th 2010, 11:15 AM
Jhevon
Quote:

Originally Posted by oblixps
i know how to do basic proofs, but some proofs on the actual limit theorems confuse me. my textbook's choices for delta are very obscure and i have no idea how they even came up with that.

for the proof of the limit theorem where the limit of a product of 2 functions is equal to the product of their limits, my book did: f = L1 + (f-L1) and g = L2 + (g-L2). and they want to show that |fg - L1*L2| < ε if 0<|x-a|<δ.

so with substitution and rearrangement they get |L1(g-L2)+L2(f-L1)+(f-L1)(g-L2)|< ε. since the limit of f as x approaches a is L1 and limit of g as x approaches a is L2, we can find positive numbers δ1, δ2, δ3, δ4 such that:

|f-L1|< sqrt(ε/3) if 0<|x-a|<δ1
|f-L1|< ε/[3(1+|L2|)] if 0<|x-a|<δ2
|g-L2|< sqrt(ε/3) if 0<|x-a|<δ3
|g-L2|< ε/[3(1+|L1|)] if 0<|x-a|<δ4

the rest of the proof i understand but what confuses me is where and how did they get those expressions sqrt(ε/3) and ε/[3(1+|L2|)]?

Notice that you have $|fg - L_1L_2| = |L_1(g - L_2) + L_2(f - L_1) + (f - L_1)(g - L_2)|$ $\le |L_1||g - L_2| + |L_2||f-L_1| + |(f - L_1)||(g - L_2)|$ by the triangle inequality.

Now, we want this expression to be less than $\epsilon$. Since we have three terms, it suffices to make sure each term is less than $\frac \epsilon 3$, so that their sum would be less than $\epsilon$. Hence we find that:

$|L_1||g - L_2| < \frac \epsilon 3 \implies |g - L_2| < \frac \epsilon {3|L_1|}$ ..........(1)

$|L_2||f - L_1| < \frac \epsilon 3 \implies |f - L_1| < \frac \epsilon {3|L_2|}$ ..........(2)

And it seems straight-forward to make $|f - L_1| < \sqrt{ \frac \epsilon 3}$ and $|g - L_2| < \sqrt{ \frac \epsilon 3}$, so that $|f - L_1||g- L_2| < \sqrt{ \frac \epsilon 3} \sqrt{ \frac \epsilon 3} = \frac \epsilon 3$.

But there is a problem with (1) and (2), what if $L_1$ or $L_2$ are zero? Big trouble. So we can throw in an extra 1 in there to ensure that division by zero never happens. At the same time, increasing the denominator makes the fraction smaller, and keeps the expression positive, thus it is good all around for an upper bound. Got it?
• Jan 4th 2010, 08:46 PM
oblixps
thanks! my textbook completely skipped those steps you just explained and just wrote those 4 inequalities right off the bat.