Let [t] denote the greatest integer ≤ t . If λεR - \ 0, 1\ , _ x →...

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

Let's find left-hand and right-hand limit.

For left-hand limit

$x = 0 - h, x \to 0, h \to 0, h > 0$

$LHL = \lim_{h \to 0}\left|\dfrac{1 - \left(- h\right) + \left|- h\right|}{\lambda - \left(- h\right) + \left[- h\right]}\right|$

$⇒ LHL = \lim_{h \to 0}\left|\dfrac{1 + h - \left(- h\right)}{\lambda + h + \left[- h\right]}\right|$

Since $h > 0 ⇒ \left[- h\right] = - 1$

$⇒ LHL = \lim_{h \to 0}\left|\dfrac{1 + h + h}{\lambda + h - 1}\right| = \dfrac{1}{\left|\lambda - 1\right|}$

For right-hand limit

$x = 0 + h, x \to 0, h \to 0, h > 0$

$RHL = \lim_{h \to 0}\left|\dfrac{1 - h + \left|h\right|}{\lambda - h + \left[h\right]}\right|$

Since $h > 0 ⇒ \left[h\right] = 0$

$⇒ RHL = \lim_{h \to 0}\left|\dfrac{1 + h + h}{\lambda + h + 0}\right| = \dfrac{1}{\left|\lambda\right|}$

For existence of limit $LHL = RHL$

$⇒ \dfrac{1}{\left|\lambda - 1\right|} = \dfrac{1}{\left|\lambda\right|}$

Cross multiply and square both side.

$⇒ {\lambda}^{2} = {\left(\lambda - 1\right)}^{2} ⇒ {\lambda}^{2} = {\lambda}^{2} + 1 - 2\lambda$

$⇒ \lambda = \dfrac{1}{2}$

Limit $L = \dfrac{1}{\dfrac{1}{2}} = 2$