Limit contradiction in L'Hopitals Rule and Special Trig limits While in Calculus the other day I stumbled upon a contradiction in L'Hopitals Rule vs. Special Trig Limits. 
The problem looks like this 
$$
\lim_{x\to 0} \frac{\tan(x)−x}{x^3}
$$
Using L'Hopitals rule (because the limit = 0/0) 
$$
\lim_{x\to 0} \frac{\tan(x)−x}{x^3} = \lim_{x\to 0} \frac{\frac{d}{dx}(\tan(x)−x)}{\frac{d}{dx}(x^3)} = \lim_{x\to 0} \frac{\sec(x)^2−1}{3x^2}
$$
Again using L'Hopitals rule (because the limit = 0/0)
$$
\lim_{x\to 0} \frac{\sec(x)^2−1}{3x^2} = \lim_{x\to 0} \frac{\frac{d}{dx}\sec(x)^2−1}{\frac{d}{dx}3x^2} = \lim_{x\to 0} \frac{2\sec(x)^2\tan(x)}{6x} = \lim_{x\to 0} \frac{\sec(x)^2\tan(x)}{3x}
$$
Again using L'Hopitals rule (because the limit = 0/0)
$$
\lim_{x\to 0} \frac{\sec(x)^2\tan(x)}{3x} = \lim_{x\to 0} \frac{\frac{d}{dx}\sec(x)^2\tan(x)}{\frac{d}{dx}3x} = \lim_{x\to 0} \frac{4\sec(x)^2\tan(x)^2 + 2\sec(x)^4}{6} = \frac{1}{3}
$$
Now, using special trig limits...
$$
\lim_{x\to 0} \frac{\tan(x)−x}{x^3} = \lim_{x\to 0} \frac{\tan(x)}{x^3}-\frac{x}{x^3} = \lim_{x\to 0} \frac{\tan(x)}{x}*\frac{1}{x^2}-\frac{1}{x^2}
$$
With knowledge of $\tan(x)$ trig limit
$$
\lim_{x\to 0} \frac{\tan(x)}{x} = 1
$$
You can now simplify.
$$
(\lim_{x\to 0} \frac{\tan(x)}{x}*\frac{1}{x^2}-\frac{1}{x^2}) = (\lim_{x\to 0} \frac{\tan(x)}{x}*\lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2}) = (1 * \lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2})
$$
Now the end result
$$ 
\lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2} = \lim_{x\to 0}\frac{1}{x^2}-\frac{1}{x^2} = 0
$$
1/3 does not = 0. Therefore there is a discrepancy through special trig limits and L'Hopitals rule. There is the same anomaly with
$$
\lim_{x\to0}\frac{\sin(x)-x}{x^3} = (\text{through L'Hopitals}) -\frac{1}{6} \text{ or (through special trig limits) } 0
$$
On a calculator in graph or table mode as you approach 0 it seems to be 1/3.
But as it turns out, if you get very precise, the limit actually seems to be approaching 0.
This happens the same with the graph, it seems to be parabolically approaching 1/3 but if you zoom in to an extreme you see it actually approaches zero.
This original work is my own as published on Sunday, October 9, 2016 at 9:51pm. I claim all knowledge credits and fallacies that may come with this discrepancy. But overall please, prove or disprove this, or at least explain why this occurs. Thank you for your time.
(I will try to get a picture of the graph and table)
 A: Your error occurs at this step,
"You can now simplify 
$(\lim_{x\to 0} \frac{tan(x)}{x}*\frac{1}{x^2}-\frac{1}{x^2}) = (\lim_{x\to 0} \frac{tan(x)}{x}*\lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2}) = (1 * \lim_{x\to 0}\frac{1}{x^2}-\lim_{x\to 0}\frac{1}{x^2})$
In particular, this equality 
$(\lim_{x\to 0} \frac{tan(x)}{x}*\frac{1}{x^2} = \lim_{x\to 0} \frac{tan(x)}{x}*\lim_{x\to 0}\frac{1}{x^2}$. 
You can only split the products when both limits are finite. $\lim_{x\to 0}\frac{1}{x^2} = \infty$
A: The error you make can be reproduced in a simpler context: let $f(x)=x+x^2$; then
$$
\lim_{x\to0}\frac{f(x)}{x}=1
$$
With the same argument you used, we should have
$$
\lim_{x\to0}\frac{f(x)-x}{x^2}=
\lim_{x\to0}\left(\frac{f(x)}{x}\cdot\frac{1}{x}-\frac{1}{x}\right)
$$
and, since $\lim_{x\to0}f(x)/x=1$ we should get $0$. Wrong!
Indeed
$$
\lim_{x\to0}\frac{f(x)-x}{x^2}=\lim_{x\to0}\frac{x^2}{x^2}=1
$$
There are many similar cases:
$$
\lim_{x\to0}\frac{x-\sin x}{x^3}=
\lim_{x\to0}\left(\frac{1}{x^2}-\frac{\sin x}{x}\frac{1}{x^2}\right)
$$
does not allow to substitute $1$ for $\frac{\sin x}{x}$.
As you can see, this has nothing to do with l'Hôpital's rule.
You can “remove” a global factor having limit $1$ without modifying the limit; for instance,
$$
\lim_{x\to0}\frac{\sin(\sin x)-\sin x}{x^3}=
\lim_{x\to0}\frac{\sin(\sin x)-\sin x}{\sin^3x}
\underbrace{\frac{\sin^3 x}{x^3\mathstrut}}_{(*)}=
\lim_{x\to0}\frac{\sin(\sin x)-\sin x}{\sin^3x}
$$
because the part marked with $(*)$ has limit $1$. More precisely, the limit we started with exists if and only if the final one exists and, in this case, they are equal.
You may not remove such a factor if it only multiplies a summand in the whole thing, just like we can't remove $\frac{f(x)}{x}$ from the example limit above. And you should be able to see the reason why. If $\frac{f(x)}{x}$ was a “global multiplier”, there would be no problem in removing it. But $f(x)-x=x^2$, not zero. And this remainder is what counts in
$$
\lim_{x\to0}\frac{f(x)-x}{x^2}
$$
exactly like $\tan x-x$ is what counts in
$$
\lim_{x\to0}\frac{\tan x-x}{x^3}
$$

By the way, if you remember that
$$
D\tan x=\frac{1}{\cos^2x}=1+\tan^2x
$$
your application of l'Hôpital would end after the first step:
$$
\lim_{x\to0}\frac{\tan x-x}{x^3}=
\lim_{x\to0}\frac{1+\tan^2x-1}{3x^2}=
\lim_{x\to0}\frac{1}{3}\left(\frac{\tan x}{x}\right)^{\!2}
$$
