ahmet Posted June 9, 2022 Posted June 9, 2022 (edited) hi,I need the answers of quuestions given below (one of my colleague asks, I could not find at its first look line) 1) limx→∞(2−x+1)x 2) the graph for 2xx2+1 sorry I could not set the formula in the new version of this website. 1) lim x to infinity (2^-x+1)^x 2) the graph for 2x / (x^2+1) Edited June 9, 2022 by ahmet
ALine Posted June 9, 2022 Posted June 9, 2022 Theres a great graphing tool online called "Desmos" which would allow for you to make such graphs. As for ze limit there are also other online resources you can use to solve it. We can't do ze work for ya. Only provide examples tools and give assistance in conceptual issues and check process ta help ya learn.
ahmet Posted June 9, 2022 Author Posted June 9, 2022 (edited) Aline I did not understand what you said well , what does ze and ya mean? note please I am mathematics teacher and the asker of this question is another maths teacher. this is not a homework help request. in the second question ,the asker is probably asking the graph with quick analysis (i.e. local minimum/maximum points , asymptotes etc) or I suppose so. thanks Edited June 9, 2022 by ahmet
joigus Posted June 9, 2022 Posted June 9, 2022 This is what you want: https://en.wikipedia.org/wiki/Logarithmic_differentiation#:~:text=In calculus%2C logarithmic differentiation or,rather than the function itself. Plus changing exponential bases: 2^(-x)=e^(-xln2) 1
ahmet Posted June 10, 2022 Author Posted June 10, 2022 7 minutes ago, joigus said: This is what you want: https://en.wikipedia.org/wiki/Logarithmic_differentiation#:~:text=In calculus%2C logarithmic differentiation or,rather than the function itself. Plus changing exponential bases: 2^(-x)=e^(-xln2) Hi, what is your result if this works
joigus Posted June 10, 2022 Posted June 10, 2022 Sorry. I did the derivative. You need the limit. The limit is of the kind 1^(infinity), which are solved by relating them to the definition of e=lim(1+f)^(1/f) if f-> 0 when x->infinity. The result is 1, if I'm not mistaken. I'll take another look tomorrow. Tell me what you get.
ahmet Posted June 10, 2022 Author Posted June 10, 2022 (edited) 3 hours ago, joigus said: Sorry. I did the derivative. You need the limit. The limit is of the kind 1^(infinity), which are solved by relating them to the definition of e=lim(1+f)^(1/f) if f-> 0 when x->infinity. The result is 1, if I'm not mistaken. I'll take another look tomorrow. Tell me what you get. I tried this, unfortunately i could not reach a meaningful result because of the persistence of ambiguity Edited June 10, 2022 by ahmet
ahmet Posted June 10, 2022 Author Posted June 10, 2022 Hi once again the second question resolved, the first one is also resolved but i could not be sure for that one.
joigus Posted June 10, 2022 Posted June 10, 2022 (edited) Hi. I did not address the second question, because I figured you would solve it as soon as you applied yourself to it. Apparently you've solved the 1st one but are unsure of its validity? The first question is best addressed with assymptotics. Here's a sketch: \[\left(1+2^{-x}\right)^{x}\sim\left(1+2^{-x}\right)^{2^{x}x2^{-x}}\] Now, \[\left(1+2^{-x}\right)^{2^{x}}\sim e\] And, \[x2^{-x}\sim0\] So, \[\left(1+2^{-x}\right)^{x}\sim e^{0}=1\] Now, making it rigorous is another matter. In fact, I went dangerously off-bounds when I said \(x2^{-x}\sim0\), as you're not supposed to ever say anything is ∼0 in assymptotics. Edited June 10, 2022 by joigus correcting for bad LateX rendering
ahmet Posted June 10, 2022 Author Posted June 10, 2022 2 hours ago, joigus said: Hi. I did not address the second question, because I figured you would solve it as soon as you applied yourself to it. Apparently you've solved the 1st one but are unsure of its validity? The first question is best addressed with assymptotics. Here's a sketch: (1+2−x)x∼(1+2−x)2xx2−x Now, (1+2−x)2x∼e And, x2−x∼0 So, (1+2−x)x∼e0=1 Now, making it rigorous is another matter. In fact, I went dangerously off-bounds when I said x2−x∼0 , as you're not supposed to ever say anything is ∼0 in assymptotics. really , I tried to benefit from (1+1/n)^n=e (n--->infinity) but reached to the same result and as you mention I could not be sufficiently sure for the limit appearing in the last line.
joigus Posted June 10, 2022 Posted June 10, 2022 Is what you want anywhere at the level of rigorousness as proving that, \[ \forall\varepsilon>0\:\exists N>0\:/\:x>N\Rightarrow\left|\left(1+2^{-x}\right)^{x}-1\right|<\varepsilon \] ?
ahmet Posted June 10, 2022 Author Posted June 10, 2022 (edited) 28 minutes ago, joigus said: Is what you want anywhere at the level of rigorousness as proving that, ∀ε>0∃N>0/x>N⇒∣∣(1+2−x)x−1∣∣<ε Ah yes this was good! but... shall we be able to find the delta correlated to epsilon here? Edited June 10, 2022 by ahmet
joigus Posted June 10, 2022 Posted June 10, 2022 Wouldn't you be satisfied with the regular lemmas and theorems, like L'Hôpital and such? Finding an N(epsilon) is a pain in the neck. 1
ahmet Posted June 10, 2022 Author Posted June 10, 2022 27 minutes ago, joigus said: Wouldn't you be satisfied with the regular lemmas and theorems, like L'Hôpital and such? Finding an N(epsilon) is a pain in the neck. hahahaha haha ha ha
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