calculus - 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)