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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. –  Ross Millikan Aug 30 '11 at 23:58
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up vote 1 down vote accepted

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. –  Joe Aug 31 '11 at 0:06
    
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. –  Doug Spoonwood Aug 31 '11 at 0:17
    
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. –  André Nicolas Aug 31 '11 at 2:08
    
While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. –  Arkamis Aug 31 '12 at 21:17
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