Without using l'hopital's rule 1

$\begin{aligned} L & = \lim_{x \to 0} \frac {1-{\color{#3D99F6}\cos x}}x & \small \color{#3D99F6} \text{By Maclaurin series} \\ & = \lim_{x \to 0} \frac {1-{\color{#3D99F6}\left(1-\frac {x^2}{2!} +\frac {x^4}{4!}- \frac {x^6}{6!}+...\right)}}x \\ & = \lim_{x \to 0} \frac {\frac {x^2}{2!} -\frac {x^4}{4!}+ \frac {x^6}{6!}+...}x \\ & = \lim_{x \to 0} \left( \frac {x}{2!} -\frac {x^3}{4!}+ \frac {x^5}{6!}+... \right) \\ & = \boxed{0} \end{aligned}$

$\displaystyle \lim_{x\to0} \dfrac{1-\cos x}x \\ = \displaystyle \lim_{x\to0} \dfrac{(1-\cos x)(1+\cos x)}{x(1+\cos x)} \\ = \displaystyle \lim_{x\to0} \dfrac{\sin^2 x}{x(1+\cos x)} \\ = \displaystyle \lim_{x\to0} \dfrac{\sin x}x \cdot \dfrac{\sin x}{(1+\cos x)} \\ = 1 \times 0 = \space \boxed{0}$ .

another solution is to graph (1-cosx)/(x) and you will find that it's approaching 0 from positive and negative direction

$\begin{aligned}\large \lim_{x\to0} \dfrac{1-\cos x}x &=\large \lim_{x\to0}\dfrac{2\sin^2(\dfrac{x}{2})}x\\ &=\large \lim_{x\to0}\dfrac{x}{2}\times\large \lim_{x\to0}\left(\dfrac{\sin(\dfrac{x}{2})}{\dfrac{x}{2}}\right)^2\\ &=0\times1=0\end{aligned}$

x趋向于0时cosx趋向于1,1-cosx则无限趋近于0，在分母x无限趋近于0却不为0时分式有意义，所以L可以无限趋近于0，但严格地说不等于0

Since $x$ appoaches $0$ , we can assign a value that is very near to $0$ , then substitute and use a calculator. Let $x=0.00001$

$\lim_{x\to0}\dfrac{1-\cos x}x=\dfrac{1-\cos 0.00001}{0.00001}=0$