Thursday, April 21, 2016

calculus - Are all limits solvable without L'Hôpital Rule or Series Expansion

Is it always possible to find the limit of a function without using L'Hôpital 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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