로피탈 정리

If we plug in $x=0$, both numerator and denominator become zero. So, we have exactly one of the indeterminate forms that l'Hopital's Rule applies to.

$$\begin{aligned} \lim_{x\rightarrow 0}\underbrace{\dfrac{\sin(x^3+3x^2)}{\sin^2x}}_{\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to 0}} &= \lim_{x\rightarrow 0}\dfrac{(3x^2+6x)\cos(x^3+3x^2)}{2\sin x\cos x} \\ \text{If we plug in $x=0$, still we find that both the numerator and the denominator go to zero. We could jump in with another iteration of l'H\^opital's Rule. However, the derivatives would be a little messy, so we use limit laws and break up the fraction into the product of two fractions. If both limits exist:} \\ \lim_{x\rightarrow 0}\dfrac{(3x^2+6x)\cos(x^3+3x^2)}{2\sin x\cos x}&=\left(\lim_{x\rightarrow 0}\dfrac{x^2+2x}{\sin x}\right)\cdot \left(\lim_{x\rightarrow 0}\dfrac{3\cos(x^3+3x^2)}{2\cos x}\right) \\ \text{We can evaluate the right-hand limit by simply plugging in $x=0$:} \\ &=\dfrac{3}{2}\lim_{x\rightarrow 0}\underbrace{\dfrac{x^2+2x}{\sin x}}_ {\genfrac{}{}{0in}{}{\mathrm{num}\to0}{\mathrm{den}\to 0}} \\ &= \dfrac{3}{2}\lim_{x\rightarrow 0}\dfrac{2x+2}{\cos x}\\ &=\frac{3}{2}\left(\frac{2}{1}\right)=3\end{aligned}$$