Prove that the limit of (a,b)→(0,0)  of f(a,b) equals 0. Use the definition of a limit. (The comparison of δ and ε) The function is

f(a,b)=(a+b)  / (6/4+cos(a))

So far I am here:

Using the inequality (a+b)^2 ≤ 2^2(a^2+b^2), we have

|a+b| ≤ 2 ∥f(a,b)∥

given ε>0  find δ>0

I got this problem some time ago. At this point I just want to know how its done.

Let a=δ, b=ε, so f(δ,ε)=(δ+ ε)/(6/(4+1) because cos(ε)→1 when ε is very small.

δ+ ε→0 when δ and ε are both small, so f(δ,ε)→0/(1.2)=0.

Therefore in the limit as (a,b)→0, f(a,b)→0.

(a+b)2=a2+b2+2ab>a2+b2 when a, b are both positive, or both negative, because 2ab>0. Therefore, (a+b)2>a2+b2, or a2+b2<(a+b)2.

When either a or b (but not both) is negative, (a+b)2=a2+b2-2ab<a2+b2. Therefore, (a+b)2<a2+b2.

a2+b2>0 for all a,b. So 3(a2+b2)>0, and, adding a2+bto both sides:

4(a2+b2)>a2+b2 or a2+b2<22(a2+b2), a2+b2+2ab<22(a2+b2)+2ab, that is:

(a+b)2<22(a2+b2)+2ab⇒(a+b)2<22(a2+b2), because 2ab>0, when a,b>0. Only when a=b=0, (a+b)2=0, 2ab=0, a2+b2=0, so (a+b)2=22(a2+b2)+2ab.

It follows then that for a,b≥0, (a+b)2≤22(a2+b2), since 22(a2+b2)≤22(a2+b2)+2ab.

f(a,b)=(a+b)/(6/4+cos(a)))=(a+b)(4+cos(a))/6. Note that since cosine is always between -1 and 1, 4+cos(a) can never be negative. 2f(a,b)=(a+b)(4+cos(a))/3, and 2|f(a,b)|=2|a+b|/(4+cos(a))/3.

If |a+b|≤2|f(a,b)|, that is, if |a+b|≤|a+b|(4+cos(a))/3, then 4+cos(a)≥3, cos(a)≥-1, which is true for all a because -1≤cos(a)≤1. So, if a=δ, δ, too, is unrestricted. This applies whatever b is because the term |a+b| cancels out provided a+b≠0. Since we are given that ε>0, a+b will never be zero.

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