Question

l'Hopital not allowed

Answer 1

I'Hopital?..

Answer 2

\[\lim_{x\rightarrow 0} \frac{tanx - x}{x^3}\]


[maybe i just don't know how to do a circumflex in latex]

Answer 3

hint:
we can use the Taylor expansion, around x=0, for tan x function, so we can write this:
\[\Large \begin{gathered}
\tan x = x + \frac{{{x^3}}}{3} + o\left( {{x^4}} \right) \Rightarrow \tan x - x = \frac{{{x^3}}}{3} + o\left( {{x^4}} \right) \hfill \\
\hfill \\
\frac{{\tan x - x}}{{{x^3}}} = \frac{{\frac{{{x^3}}}{3} + o\left( {{x^4}} \right)}}{{{x^3}}} \to \frac{1}{3} \hfill \\
\end{gathered} \]

Answer 4

`\hat{...}` for all your circumflex needs :)

e.g. `\hat{f}` gives \(\large\hat{f}\), `\text{L'Hopit}\hat{\text{a}}\text{l}` gives \(\text{L'Hopit}\hat{\text{a}}\text{l}\)

Answer 5

Another approach, which is only valid provided you can prove the limit exists. Assume the limit is \(L\). Then
\[L=\lim_{x\to0}\frac{\tan x-x}{x^3}=\lim_{x\to0}\left(\frac{\tan x}{x^3}-\frac{1}{x^2}\right)\]
This next part is a bit hand-wavy, 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:
\[\begin{align*}
L&=\lim_{x\to0}\left(\frac{\tan 2x}{(2x)^3}-\frac{1}{(2x)^2}\right)\\[1ex]
8L&=\lim_{x\to0}\left(\frac{\tan 2x}{x^3}-\frac{2}{x^2}\right)\\[1ex]
&=\lim_{x\to0}\left(\frac{\frac{2\tan x}{1-\tan^2x}}{x^3}-\frac{2}{x^2}\right)\\[1ex]
4L&=\lim_{x\to0}\left(\frac{\tan x}{x^3(1-\tan^2x)}-\frac{1}{x^2}\right)\\[1ex]
4L-L&=\lim_{x\to0}\left(\frac{\tan x}{x^3(1-\tan^2x)}-\frac{1}{x^2}\right)-\lim_{x\to0}\left(\frac{\tan x}{x^3}-\frac{1}{x^2}\right)\\[1ex]
3L&=\lim_{x\to0}\left(\frac{\tan x}{x^3(1-\tan^2x)}-\frac{\tan x}{x^3}\right)\\[1ex]
&=\color{red}{\lim_{x\to0}\frac{\tan x}{x}}\lim_{x\to0}\left(\frac{1}{x^2(1-\tan^2x)}-\frac{1}{x^2}\right)\\[1ex]
&=\color{red}1\times\lim_{x\to0}\frac{\tan^2x}{x^2(1-\tan^2x)}\\[1ex]
&=\color{red}{\lim_{x\to0}\frac{\tan^2x}{x^2}}\lim_{x\to0}\frac{1}{1-\tan^2x}\\[1ex]
&=1\\[1ex]
L&=\frac{1}{3}
\end{align*}\]

Answer 6

cool stuff!!

and absolutely no need for the Hospital

Answer 7

\[\text{L'H} \hat{ \text{o}} \text{pital}\]

Answer 8

@SithsAndGiggles Would this be sufficient?
\[\lim_{x\to 0}f(kx)=\lim_{kx\to 0}f(kx)=\lim_{X\to0}f(X)=\lim_{x\to 0}f(x)\]

Answer 9

@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...