Limit of a function using the calculator The problem is: find $\lim_{x \to 0}\frac{\tan(x) - x}{x^3}$ using the calculator.
I know that the limit should be $\frac{1}{3}$, but what I get plugging in very small numbers is $0$.
Now, I know that the function is oscillating very fast about the origin, but still I was expecting to find good approximations of 0.33333 taking smaller and smaller $x$...I was expecting this since this reflects my idea of limit...
Could you explain me what is going on?
 A: You are suffering from loss of significance.  You are considering that $\tan (x) \approx x + \frac {x^3}3$ for $x \ll 1$  which is correct.  Your calculator stores numbers with (say) $10$ decimal places.  If you ask for $\tan(10^{-10})-10^{-10}$  it will calculate each term to $10$ place accuracy, but they are equal.   $\frac {x^3}3=\frac 13\cdot 10^{-30}$ so the calculator doesn't know about it.  The numerator is then zero and you get zero.  As noted in the comments, if you take $x$ somewhat larger, like $0.1$ or $0.001$ you will get something close to $\frac 13$
A: Actually, there is no oscillation near the origin (things like $\sin(1/x)$, e.g. will oscillate rapidly near the origin, but $\lim_{x\to 0} \tan(x) = 0$ is pretty straightforward).
I would check carefully the expression you have entered in the calculator. For instance, if you accidentally entered something like

tan(x - x)/x^3

etc.
Other answers have suggested a more fitting culprit for this problem, but I'll stand by my claim that the function does not oscillate near the origin.
A: Your idea of a limit is perfecly right, but the calculation of tan(x)-x for very small
 x is numerically instable (You subtract two nearly equal numbers).
This is the reason, that you got 0. In theory, the approximation gets better and
 better.
