Let L 1-lim x - 0cos - x e - x -1-sin x and L 2-lim x - 0ln 1- x -sin 2 x - x- If L 1- L 2- then the value of - isNote- - denotes the largest integer less than or equal to -

L_1 =lim_x → 0cos (π x) (e^λ x 1)πsin x ⇒ nbsp;L_1 = lim_x → 0cos (π x) (e^λ x 1) λπ (λ x)sin xx ⇒ L_1 =λπ ⋯(1) L_2 = lim_x → 0ln (1 x) +sin 2xx ⇒ L_2 = li ...

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Byju's Answer

Standard XII

Mathematics

Standard Limits to Remove Indeterminate Form

Let L 1=lim x...

Question

Let $L_{1} = lim x \to 0 \frac{cos (π x) (e^{λ x} - 1)}{π sin x}$ and $L_{2} = lim x \to 0 \frac{ln (1 - x) + sin 2 x}{x}$ . If $L_{1} = L_{2}$ , then the value of $[λ]$ is
(Note: $[λ]$ denotes the largest integer less than or equal to $λ$ .)

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Solution

$L_{1} = lim x \to 0 \frac{cos (π x) (e^{λ x} - 1)}{π sin x} \Rightarrow L_{1} = lim x \to 0 \frac{cos (π x) (e^{λ x} - 1) λ}{π (λ x) \frac{sin x}{x}} \Rightarrow L_{1} = \frac{λ}{π} \dots (1) L_{2} = lim x \to 0 \frac{ln (1 - x) + sin 2 x}{x} \Rightarrow L_{2} = lim x \to 0 \frac{ln (1 - x)}{x} + lim x \to 0 \frac{2 sin 2 x}{2 x} \Rightarrow L_{2} = - 1 + 2 = 1 \dots (2)$
We know that,
$L_{1} = L_{2}$
Using equation $(1)$ and $(2)$ ,
$\Rightarrow \frac{λ}{π} = 1 \Rightarrow λ = π \Rightarrow [λ] = 3$

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Let L1=limx→0cos(πx)(eλx−1)πsinx and L2=limx→0ln(1−x)+sin2xx. If L1=L2, then the value of [λ] is (Note: [λ] denotes the largest integer less than or equal to λ.)

L1=limx→0cos(πx)(eλx−1)πsinx⇒L1=limx→0cos(πx)(eλx−1)λπ(λx)sinxx⇒L1=λπ ⋯(1)L2=limx→0ln(1−x)+sin2xx⇒L2=limx→0ln(1−x)x+limx→02sin2x2x⇒L2=−1+2=1 ⋯(2) We know that, L1=L2 Using equation (1) and (2), ⇒λπ=1⇒λ=π⇒[λ]=3

Let $L_{1} = lim x \to 0 \frac{cos (π x) (e^{λ x} - 1)}{π sin x}$ and $L_{2} = lim x \to 0 \frac{ln (1 - x) + sin 2 x}{x}$ . If $L_{1} = L_{2}$ , then the value of $[λ]$ is
(Note: $[λ]$ denotes the largest integer less than or equal to $λ$ .)