# Are all limits solvable without L'Hospital Rule or Series Expansion

Is it always possible to find the limit of a function without using L'Hospital Rule or Series Expansion

For example,

$$\lim_{x\to0}\frac{\tan x-x}{x^3}$$

$$\lim_{x\to0}\frac{\sin x-x}{x^3}$$

$$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$

$$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$

$$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$

$$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$

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In general, $\lim_{x \to 0} \frac{f(x) - \sum_{k = 1}^{n - 1} \frac{f^{(k)}(0)\cdot x^k}{k!}}{x^n} = \frac{f^{(n)}(0)}{n!}$. This can be proven using the Mean Value Theorem $n$ times and induction.
What I mean is that this case of l'Hôpital's theorem is easily proved using the mean value theorem. So, applying this case can really be "without l'Hôpital or Taylor expansion"? Your assertion is mostly the same as Taylor expansion, I believe. Anyway, the question is not well posed. To me, applying $\lim_{x\to0}(\sin x)/x=1$ is just the same as using the derivative of $\sin$, so l'Hôpital or Taylor. –  egreg May 13 '13 at 23:19
No, that is circular logic. You need that limit before you can even compute $\frac{d}{dx} \sin x$, which is needed for all three things you listed. It is most certainly not that same thing; in fact, it's not even a subject of calculus so much as precalculus and basic limits. –  Jon Claus May 13 '13 at 23:48