Let [t] denote the greatest integer leq t. If | vedclass

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(B) For the limit to exist,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$.$LHL = \lim_{x \rightarrow 0^{-}} \left| \frac{1-x+(-x)

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$2$

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(B) For the limit to exist,the left-hand limit $(LHL)$ must equal the right-hand limit $(RHL)$.$LHL = \lim_{x \rightarrow 0^{-}} \left| \frac{1-x+(-x)}{\lambda-x+(-1)} \right| = \left| \frac{1}{\lambda-1} \right|$$RHL = \lim_{x \rightarrow 0^{+}} \left| \frac{1-x+x}{\lambda-x+0} \right| = \left| \frac{1}{\lambda} \right|$Equating $LHL$ and $RHL$:$\left| \frac{1}{\lambda-1} \right| = \left| \frac{1}{\lambda} \right| \Rightarrow |\lambda| = |\lambda-1|$Squaring both sides: $\lambda^2 = \lambda^2 - 2\lambda + 1$ $\Rightarrow 2\lambda = 1$ $\Rightarrow \lambda = \frac{1}{2}$.Substituting $\lambda = \frac{1}{2}$ into the expression for $L$:$L = \left| \frac{1}{1/2} \right| = 2$.

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

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