calculus - Evaluate $lim_{xto0}frac{1-cos3x+sin 3x}x$ without L'Hôpital's rule

I've been trying to solve this question for hours. It asks to find the limit without L'Hôpital's rule. $$\lim_{x\to0}\frac{1-\cos3x+\sin3x}x$$ Any tips or help would be much appreciated.

Answer

If you are given that $\lim_{x \to 0}{\sin x \over x } = 1$, then since $1-\cos (3x) = 2 \sin^2 ({3 \over 2} x)$ (half angle formula), we have

$$\begin{eqnarray} {1 -\cos (3x) +\sin (3x) \over x } &=& {2 \sin^2 ({3 \over 2} x) \over x} +{\sin (3x) \over 3x} {3x \over x} \\ &=& 2 ({\sin ({3 \over 2} x) \over {3 \over 2} x})^2 { ({3 \over 2} x)^2 \over x} + 3 {\sin (3x) \over 3x} \\ &=& {9 \over 2} x ({\sin ({3 \over 2} x) \over {3 \over 2} x})^2 + 3 {\sin (3x) \over 3x} \end{eqnarray}$$ Taking limits gives $3$.