Evaluate $\lim_{x\to0}\frac{1-\cos3x+\sin 3x}x$ 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.

Taylor expansion is always a good solution since the method will provide the limit and more.
Remembering that $$\cos(t)=1-\frac{t^2}{2}+O\left(t^4\right)$$ $$\sin(t)=t-\frac{t^3}{6}+O\left(t^4\right)$$ replace $t$ by $3x$ to get $$1-\cos (3 x)+\sin (3 x)=1-\left(1-\frac{9 x^2}{2}+O\left(x^4\right) \right) +\left( 3 x-\frac{9 x^3}{2}+O\left(x^4\right)\right)$$ $$1-\cos (3 x)+\sin (3 x)=3 x+\frac{9 x^2}{2}-\frac{9 x^3}{2}+O\left(x^4\right)=3 x+\frac{9 x^2}{2}+O\left(x^3\right)$$ $$\frac{1-\cos (3 x)+\sin (3 x) }x=3+\frac{9 x}{2}+O\left(x^2\right)$$ which shows the limit and how it is approached.
