Let $$[t]$$ denote the greatest integer $$\leq t$$. If $$\lambda \in R - \{0, 1\}$$, $$\lim_{x \to 0}\left|\frac{1 - x + |x|}{\lambda - x + [x]}\right| = L$$, then $$L$$ is equal to

We are given $$[t]$$ denotes the greatest integer $$\leq t$$. We need to find:

$$L = \lim_{x \to 0} \left|\frac{1 - x + |x|}{\lambda - x + [x]}\right|$$

where the outer bars denote absolute value, the numerator contains $$|x|$$ (absolute value), and the denominator contains $$[x]$$ (greatest integer function).

Case 1: $$x \to 0^+$$

When $$x > 0$$ and close to 0: $$|x| = x$$ and $$[x] = 0$$.

Numerator $$= 1 - x + x = 1$$

Denominator $$= \lambda - x + 0 = \lambda - x$$

So the expression $$= \left|\frac{1}{\lambda - x}\right| \to \left|\frac{1}{\lambda}\right| = \frac{1}{|\lambda|}$$ as $$x \to 0^+$$.

Case 2: $$x \to 0^-$$

When $$-1 < x < 0$$: $$|x| = -x$$ and $$[x] = -1$$.

Numerator $$= 1 - x + (-x) = 1 - 2x$$

Denominator $$= \lambda - x + (-1) = (\lambda - 1) - x$$

So the expression $$= \left|\frac{1 - 2x}{(\lambda - 1) - x}\right| \to \left|\frac{1}{\lambda - 1}\right| = \frac{1}{|\lambda - 1|}$$ as $$x \to 0^-$$.

For the limit $$L$$ to exist, both one-sided limits must be equal:

$$\frac{1}{|\lambda|} = \frac{1}{|\lambda - 1|}$$

This gives $$|\lambda| = |\lambda - 1|$$.

Squaring both sides: $$\lambda^2 = (\lambda - 1)^2$$

$$\lambda^2 = \lambda^2 - 2\lambda + 1$$

$$0 = -2\lambda + 1$$

$$\lambda = \frac{1}{2}$$

Since $$\lambda = \frac{1}{2} \in \mathbb{R} - \{0, 1\}$$, this is a valid value.

Substituting $$\lambda = \frac{1}{2}$$:

$$L = \frac{1}{|\lambda|} = \frac{1}{1/2} = 2$$

Therefore, $$L = 2$$, which corresponds to Option B.