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Any ideas how to show $\lim\limits_{x\to 0}\dfrac{x^2-1+\cos^2x}{x^4+x^3\sin x}=\lim\limits_{x\to 0}\dfrac{x-\sin(x)}{x^3}=\dfrac{1}{6}$ without using the De L'Hôpital rule (or proving a special case of it?). How can we reduce this to $\lim\limits_{x\to 0}\dfrac{1-\cos(x)}{x^2}=\dfrac{1}{2}$?

You can suppose that we know the limit in question exists and therefore use inequalities to bound it

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    Sorry, the coefficient of $sin$ is actually $x^3$2012-11-01
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    Consider taking Taylor Expansions for both non-polynomial functions and see what happens.2012-11-01
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    What exactly is your question? Do you want to know how to reduce the original limit to $\lim_{x \to 0} \frac{x - \sin x}{x^3}$? Or, do you already know that, in which case your question is just how to determine the last limit is 1/6??? If it's just the second question, then this is an EXACT duplicate of http://math.stackexchange.com/q/217081/86712012-11-01
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    I guess this question isn't an EXACT duplicate because the previous one does not say we can assume the limit exists. So, this question is in fact easier, and the answer provided in the other question actually works for this one. On the other hand, the answer provided by Hans mentioned in that one works even if you don't assume the limit exists.2012-11-01
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    Here's an idea: if you are allowed to use derivatives then you can establish polynomial bounds for the numerator and denominator (their Taylor polynomials to an $n$ and $n+1$st degree) and then use the squeeze theorem appropriately.2012-11-01

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