1. We need to evaluate $L=\lim_{x\to 0}\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|,$ where $[x]$ is the greatest integer function and $\lambda\in\mathbb R\setminus\{0,1\}$.

2. Since both $|x|$ and $[x]$ behave differently for $x\to 0^+$ and $x\to 0^-$, we check one-sided limits.

Case 1: $x\to 0^+$

For $x>0$ sufficiently small, $|x|=x, \qquad [x]=0.$ So, $1-x+|x|=1-x+x=1,$ and $\lambda-x+[x]=\lambda-x.$ Hence $\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|=\left|\frac{1}{\lambda-x}\right|.$ Therefore, $\lim_{x\to 0^+}\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|=\frac{1}{|\lambda|}.$

Case 2: $x\to 0^-$

For $x<0$ sufficiently close to $0$, $|x|=-x, \qquad [x]=-1.$ So, $1-x+|x|=1-x-x=1-2x,$ and $\lambda-x+[x]=\lambda-x-1.$ Thus, $\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|=\left|\frac{1-2x}{\lambda-1-x}\right|.$ Taking limit as $x\to 0^-$, $\lim_{x\to 0^-}\left|\frac{1-x+|x|}{\lambda-x+[x]}\right|=\left|\frac{1}{\lambda-1}\right|=\frac{1}{|\lambda-1|}.$

3. For the two-sided limit to exist, the right-hand and left-hand limits must be equal: $\frac{1}{|\lambda|}=\frac{1}{|\lambda-1|}.$ So, $|\lambda|=|\lambda-1|.$

4. Solve this equation: $\lambda^2=(\lambda-1)^2$ $\lambda^2=\lambda^2-2\lambda+1$ $2\lambda=1$ $\lambda=\frac12.$ This satisfies the condition $\lambda\ne 0,1$.

5. Now compute the common limit: $L=\frac{1}{|\lambda|}=\frac{1}{\left|\frac12\right|}=2.$

Thus, $L=2.$

6. Comparing with the options, the correct option is B.