prove the statement using epsilon delta definition of a limit lim (3-4/5x)=-5 x-

Question

prove the statement using epsilon delta definition of a limit lim (3-4/5x)=-5 x->10

Explanation / Answer

1. We know that for all eps > 0 and for all k, there is an n > k such that x_n "is really near to" 5. We are now given an eps2 and a k2. And we must prove that there is an n2 > k2 such that 4(x_n) is nearer to 20 than eps2. Let's first try to use eps=eps2 and k=k2. We start with |4x_n - 20| = 4|x_n - 5| < 4eps = 4eps2 (The k = k2). Mhm - this is 4 times too large: We want the difference on the left to be smaller than eps2. So we try again with eps=eps2/4 and k=k2. Then we get |4x_n - 20| = 4|x_n - 5| < 4eps = 4eps2/4 = eps2 Hooray! We have shown that |4x_n - 20| is actually smaller than a given eps2 for all n > k2 = k. 2. ([(3n+20)/n] x (-1)^n) ...mhm, it's the same as ([3 + 20/n] x (-1)^n). If we just take all the elements with even indices, we get the sequence (3 + 20/2n) Now, if we can prove that this pseudo-converges to 3, we are done, aren't we? Because we can then, for some eps and k, just select the next *even* n that is larger k and fulfill the condition with this one. Now, the above is equal to the sequence (3 + 10/n). We assume we are given that eps and k; and we must find the n such that |(3 + 10/n) - 3| n 5.77... > n But there is no n < 5.77 that is, at the same time, > 100 !! So for this specific eps=0.01 and k=100, we cannot find a suitable n ... hence no pseudo-convergence ... (How did I find the values for eps and k? First I tried 10^-6 and 10^6. But then I saw that I had to compute 1/25 - eps and thought "0.04 - 0.01 should be enough" and rewrote it with 0.01 and 100 ...).

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