# Finding the limit with binomials in the denominator

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

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