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IrishBoy123
 one year ago
l'Hopital not allowed
IrishBoy123
 one year ago
l'Hopital not allowed

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\rightarrow 0} \frac{tanx  x}{x^3}\] [maybe i just don't know how to do a circumflex in latex]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3hint: we can use the Taylor expansion, around x=0, for tan x function, so we can write this: \[\Large \begin{gathered} \tan x = x + \frac{{{x^3}}}{3} + o\left( {{x^4}} \right) \Rightarrow \tan x  x = \frac{{{x^3}}}{3} + o\left( {{x^4}} \right) \hfill \\ \hfill \\ \frac{{\tan x  x}}{{{x^3}}} = \frac{{\frac{{{x^3}}}{3} + o\left( {{x^4}} \right)}}{{{x^3}}} \to \frac{1}{3} \hfill \\ \end{gathered} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0`\hat{...}` for all your circumflex needs :) e.g. `\hat{f}` gives \(\large\hat{f}\), `\text{L'Hopit}\hat{\text{a}}\text{l}` gives \(\text{L'Hopit}\hat{\text{a}}\text{l}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Another approach, which is only valid provided you can prove the limit exists. Assume the limit is \(L\). Then \[L=\lim_{x\to0}\frac{\tan xx}{x^3}=\lim_{x\to0}\left(\frac{\tan x}{x^3}\frac{1}{x^2}\right)\] This next part is a bit handwavy, but it seems true at first glance (still thinking of a proof...), but I'm pretty sure that \[\lim_{x\to0}f(x)=L~~\implies~\lim_{x\to0}f(kx)=L\] for arbitrary \(k\neq0\). Now, I could be very wrong, but I'm thinking of how \(\lim\limits_{x\to0}\dfrac{\sin kx}{kx}=1\) for arbitrary \(k\neq0\). I have a good feeling this might be true. I'm going to use this "fact" and say that the following must also be true: \[L=\lim_{x\to0}\left(\frac{\tan 2x}{(2x)^3}\frac{1}{(2x)^2}\right)\] The choice of \(k=2\) is arbitrary, I could have easily used any other real number, but this one is easiest to work with: \[\begin{align*} L&=\lim_{x\to0}\left(\frac{\tan 2x}{(2x)^3}\frac{1}{(2x)^2}\right)\\[1ex] 8L&=\lim_{x\to0}\left(\frac{\tan 2x}{x^3}\frac{2}{x^2}\right)\\[1ex] &=\lim_{x\to0}\left(\frac{\frac{2\tan x}{1\tan^2x}}{x^3}\frac{2}{x^2}\right)\\[1ex] 4L&=\lim_{x\to0}\left(\frac{\tan x}{x^3(1\tan^2x)}\frac{1}{x^2}\right)\\[1ex] 4LL&=\lim_{x\to0}\left(\frac{\tan x}{x^3(1\tan^2x)}\frac{1}{x^2}\right)\lim_{x\to0}\left(\frac{\tan x}{x^3}\frac{1}{x^2}\right)\\[1ex] 3L&=\lim_{x\to0}\left(\frac{\tan x}{x^3(1\tan^2x)}\frac{\tan x}{x^3}\right)\\[1ex] &=\color{red}{\lim_{x\to0}\frac{\tan x}{x}}\lim_{x\to0}\left(\frac{1}{x^2(1\tan^2x)}\frac{1}{x^2}\right)\\[1ex] &=\color{red}1\times\lim_{x\to0}\frac{\tan^2x}{x^2(1\tan^2x)}\\[1ex] &=\color{red}{\lim_{x\to0}\frac{\tan^2x}{x^2}}\lim_{x\to0}\frac{1}{1\tan^2x}\\[1ex] &=1\\[1ex] L&=\frac{1}{3} \end{align*}\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0cool stuff!! and absolutely no need for the Hospital

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0\[\text{L'H} \hat{ \text{o}} \text{pital}\]

amilapsn
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles Would this be sufficient? \[\lim_{x\to 0}f(kx)=\lim_{kx\to 0}f(kx)=\lim_{X\to0}f(X)=\lim_{x\to 0}f(x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@amilapsn yeah that should work for the "fact". As for proving whether the limit exists, we might be able to get away with the squeeze theorem, but I'm not seeing any obvious bounds...
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