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.
 A: Do you know how to do without L'Hôpital's rule
$$
\lim_{x\to 0}\frac{\sin x}{x}
\quad \text{
and   }\quad 
\lim_{x\to 0}\frac{\cos x-1}{x}?
$$
Then try the "simpler" version of your problem:
$$
\lim_{x\to 0}\frac{\sin 3x}{x},
\qquad
\lim_{x\to 0}\frac{\cos 3x-1}{x}.
$$
A: 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 $3$.
A: Since $\lim_{x\to 0}\frac{\sin x}{x}=1$ it follows that $\lim_{x\to 0}\frac{\sin(3x)}{x}=3$.
On the other hand, since 
$$ 1-\cos(3x) = 2\sin^2\left(\frac{3x}{2}\right) $$
it follows that $\lim_{x\to 0}\frac{1-\cos(3x)}{x}=0$. It should not be difficult to finish from here.
A: If I write it in this form, does it look familiar ?
$$\lim_{x\to 0} \frac{[-\cos(3x) + \sin(3x)] - (-1)}{x - 0}$$
$$\lim_{x\to 0} \frac{[-\cos(3x) + \sin(3x)] - [-cos(0) + sin(0)]}{x - 0}$$
A: 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.
A: Here is a similar approach also relying on the standard table limit $\lim\limits_{x\to 0}\frac{\sin x}{x}=1$:
$\lim\limits_{x\to 0}\frac{1-\cos(3x)+\sin(3x)}{x}=\lim\limits_{x\to 0}\frac{\frac{(1+\sin(3x))^2-\cos^2(3x)}{x}}{1+\sin(3x)+\cos(3x)}=\lim\limits_{x\to 0}\frac{2\cdot\frac{\sin (3x)}{x}(1+\sin(3x))}{1+\sin(3x)+\cos(3x)}=3$
