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

For example,

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

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





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1 Answer 1

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.

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This is, of course, a form of l'Hôpital's theorem. –  egreg May 13 '13 at 23:06
The Mean Value Theorem does not requires l'Hôpital's rule to prove, nor vice-versa for most cases where the limit is at a real value (as opposed to infinity). –  Jon Claus May 13 '13 at 23:12
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
The method above uses repeated derivatives and essentially uses the proof technique of Taylor's series. I believe the OP wanted to have a technique which avoids differentiation altogether. But when we disable differentiation, it is almost impossible to evaluate the above limits. In some of the cases if we assume the existence of the limit then we can evaluate the limit by simple algebraic/trigonometric manipulation. But for existence we need differentiation. –  Paramanand Singh Jul 7 '13 at 5:43

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