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anonymous

  • one year ago

Question 6 of the Unit 1 exam supposes f satisfies: f(x+y) = f(x) + f(y) + x^2y + xy^2 It also supposes (A) the limit of f(x)/x as x nears 0 is 1. It says to evaluate f(0). The solutions says that if A, then (B) the limit of f(x) as x nears 0 is 0. I see that given B, f(0) = 0, but how does A get you to B? It seems more obvious that since f(0+0) = f(0) + f(0) + 0^2*0 + 0*0^2, it has to be that f(0) = f(0) + f(0), but then f(0) = 0. It is clear how to go on from here, can someone explain the connection between A and B?

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  1. anonymous
    • one year ago
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    Ah, I see.... Given lim [f(x)/x] (x->0) = 1: Then: ln(lim (x->0) [f(x)/x]) = 0. Then: lim (x->0) [ln f(x) - ln x] = 0 Then: lim (x->0) [ln f(x)] = lim (x->0) [ln x] Then: ln [lim (x->0) f(x)] = ln [lim (x->0) x] Then: e^(ln [lim (x->0) f(x)]) = e^(ln [lim (x->0) x]) Then: lim (x->0) f(x) = lim (x->0) x Since lim (x->0) x = 0, so lim (x->0) f(x) = 0. Needs most of the tricks....

  2. phi
    • one year ago
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    yes, your way works. another way is start with \[ \lim_{x\rightarrow 0} \frac{f(x)}{x} = 1 \] being pedantic we can also take the limit of the right side \[ \lim_{x\rightarrow 0} \frac{f(x)}{x} = \lim_{x\rightarrow 0}1 \] as long as x is not zero we can multiply both sides by x \[ \lim_{x\rightarrow 0} f(x) = \lim_{x\rightarrow 0}x\] as x goes to zero, write this as \[ f(0)= 0 \]

  3. anonymous
    • one year ago
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    Interesting. I knew that \[\lim_{x \rightarrow 0}c f(x) = c \lim_{x \rightarrow 0} f(x)\] (cool editor) but I hadn't suspected that c could be x. Learned another trick today....

  4. anonymous
    • one year ago
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    After some time, that is a cool approach. Thanks for the lesson. That's why I looked for this place.

  5. anonymous
    • one year ago
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    phi, I just realized that *that* wasn't what you did, but what you did was interesting too.

  6. anonymous
    • one year ago
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    The key step is: lim(x>0) x * lim(x>0) f(x)/x = lim(x>0) x * lim(x>0) 1, and so: lim(x>0) x * f(x)/x = lim(x>0) x * 1, et cetera.

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