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Another problem that my friend asked me about and I didn't know the answer to, so I thought I'd ask here since I always get back on track afterwards.

The problem is getting rid of the binomial in the denominator when finding the limit of a function in the indeterminate form. Once again, after finding the indeterminate form, I have no clue what to do. $$\lim_{x\to0}\frac{\sin x}{x^2+3x}$$

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    Intuitively, as $x \to 0, x^2 \ll 3x$ so you can ignore it.2011-08-30

2 Answers 2

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HINT $\ $ Factor it as $\rm\ \dfrac{sin\ x}{x}\ \dfrac{1}{x+3}$

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Take a trip to the hospital.

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    Some of this goes much further than what we've learned in only a few weeks of class. I don't want to go too far ahead and confuse myself more than I already am.2011-08-31
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    Fair enough. But, I don't see how Bill's answer works without L'Hopital's rule, or you just assuming lim x->0 ((sin x)/x)=1. If you're assuming ((sin x)/x)=1, you might have to do that I can understand, but it also comes as derivable via L'Hopital's rule.2011-08-31
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    For most derivations of the basic facts, L'Hospital's Rule doesn't work without Bill's answer, rather than the other way around as you have it. In beginning calculus, one shows that the limit is $1$, usually by a geometric argument, in order to show that the derivative of $\sin x$ is $\cos x$. Without knowledge of the derivative, one cannot apply L'Hospital's Rule to the problem.2011-08-31