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When I solve limits I often convert a block in a notable limit multiplying and dividing it by the same quantity. For example: $$ \lim_{x \to 0} \frac{e^{\sin(x)}-1}{x} = \lim_{x \to 0} \frac{e^{\frac{\sin(x)}{x}x}-1}{x} = \lim_{x \to 0} \frac{e^{x}-1}{x} = 1 $$

I've noticed that I can't always apply this. For example: $$ \lim_{x \to 0} \frac{x^3+9x-9 \tan(x)}{-8x^3} = \frac{1}{4}\neq \lim_{x \to 0} \frac{x^3+9x-9 \frac{\tan(x)}{x}x}{-8x^3} = \lim_{x \to 0} \frac{x^3+9x-9x}{-8x^3}= -\frac{1}{8} $$

What's wrong in that passage? What rule am I violating?

Thanks

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    Technically, you can't so blithely make the change in the first example *either*. It has to be justified in terms of the order of vanishing of $\sin(x)/x$ vs. that of $x$.2012-05-03
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    To repeat what Arturo Magidin wrote, your first derivation was incorrect, but the answer happened to be right. You can argue this way if you have full knowledge of the fine-grained behaviour of the function near $x=0$, usually through knowledge of the first few relevant terms of the series expansion.2012-05-03
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    @AndréNicolas Let me know if there is anything sloppy going on with my answer. I guess it is clear, but it is rather a delicate topic.2012-05-03

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