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:

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Limits

Let

[t]

denote the greatest integer

\leq t

. If for some

λ \in R - {0, 1}

lim_{x \to 0} \frac{1 - x + ∣ x ∣}{λ - x + [ x ]} = L

, then

L

is equal to:

Answer: (D)

Approaching from the left,

∣ x ∣ = - x

and

[x] = - 1

, so the expression tends to

\frac{1}{λ - 1}

; approaching from the right,

∣ x ∣ = x

and

[x] = 0

, so it tends to

\frac{1}{λ}

. For the limit to exist these must agree,

∣ λ - 1∣ = ∣ λ ∣

, which forces

λ = \frac{1}{2}

. The common value is then

L = \frac{1}{∣ λ ∣} = \frac{1}{1/2} = 2

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option

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