JEE Main 2025 22 January Evening Shift
Exam JEE Main
Year 2025
Date 22 January
Shift Evening Shift
Subject Mathematics
Chapter Limits JEE Main 2025 · 22 January · Evening Shift · Mathematics · Limits
If _ x arrow ∞ (( e 1-e )( 1 e -(x)/(1+x)))^x=α , then the value of...
If
lim x → ∞ ( ( e 1 − e ) ( 1 e − x 1 + x ) ) x = α \lim _{x \rightarrow \infty}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{x}{1+x}\right)\right)^x=\alpha lim x → ∞ ( ( 1 − e e ) ( e 1 − 1 + x x ) ) x = α , then the value of
log e α 1 + log e α \frac{\log _{\mathrm{e}} \alpha}{1+\log _{\mathrm{e}} \alpha} 1 + l o g e α l o g e α equals :
Options A e − 1 e^{-1} e − 1 B e 2 \mathrm{e}^2 e 2 C e − 2 e^{-2} e − 2 D e
Step-by-step solution
α = lim x → ∞ ( ( e 1 − e ) ( 1 e − x 1 + x ) ) x ( 1 ∞ form ) ∴ α = e L Where L = lim x → ∞ x ( ( e 1 − e ) ( 1 e − x 1 + x ) − 1 ) ⇒ L = lim x → ∞ ( e 1 − e ) x ( 1 e − x 1 + x − ( 1 − e e ) ) ⇒ L = e 1 − e lim x → ∞ x ( 1 − x 1 + x ) ⇒ L = e 1 − e lim x → ∞ x x + 1 ⇒ L = e 1 − e ⋅ 1 ⇒ L = e 1 − e ∴ α = e e 1 − e ⇒ log α = e 1 − e ∴ Required value = e 1 − e 1 + e 1 − e = e \begin{aligned} & \alpha=\lim _{x \rightarrow \infty}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{\mathrm{x}}{1+\mathrm{x}}\right)\right)^{\mathrm{x}}\left(1^{\infty} \text{ form }\right) \\ & \therefore \alpha=\mathrm{e}^{\mathrm{L}} \\ & \text{ Where } \mathrm{L}=\lim _{x \rightarrow \infty} \mathrm{x}\left(\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right)\left(\frac{1}{\mathrm{e}}-\frac{\mathrm{x}}{1+\mathrm{x}}\right)-1\right) \\ & \Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow \infty}\left(\frac{\mathrm{e}}{1-\mathrm{e}}\right) \mathrm{x}\left(\frac{1}{\mathrm{e}}-\frac{\mathrm{x}}{1+\mathrm{x}}-\left(\frac{1-\mathrm{e}}{\mathrm{e}}\right)\right) \\ & \Rightarrow \mathrm{L}=\frac{\mathrm{e}}{1-\mathrm{e}} \lim _{\mathrm{x} \rightarrow \infty} \mathrm{x}\left(1-\frac{\mathrm{x}}{1+\mathrm{x}}\right) \\ & \Rightarrow \mathrm{L}=\frac{\mathrm{e}}{1-\mathrm{e}} \lim _{\mathrm{x} \rightarrow \infty} \frac{\mathrm{x}}{\mathrm{x}+1} \\ & \Rightarrow \mathrm{ L}=\frac{\mathrm{e}}{1-\mathrm{e}} \cdot 1 \\ & \Rightarrow \mathrm{ L}=\frac{\mathrm{e}}{1-\mathrm{e}} \\ & \therefore \alpha=\mathrm{e}^{\frac{\mathrm{e}}{1-\mathrm{e}}} \Rightarrow \log \alpha=\frac{\mathrm{e}}{1-\mathrm{e}} \\ & \therefore \text{ Required value }=\frac{\frac{\mathrm{e}}{1-\mathrm{e}}}{1+\frac{\mathrm{e}}{1-\mathrm{e}}}=\mathrm{e}\end{aligned} α = x → ∞ lim ( ( 1 − e e ) ( e 1 − 1 + x x ) ) x ( 1 ∞ form ) ∴ α = e L Where L = x → ∞ lim x ( ( 1 − e e ) ( e 1 − 1 + x x ) − 1 ) ⇒ L = x → ∞ lim ( 1 − e e ) x ( e 1 − 1 + x x − ( e 1 − e ) ) ⇒ L = 1 − e e x → ∞ lim x ( 1 − 1 + x x ) ⇒ L = 1 − e e x → ∞ lim x + 1 x ⇒ L = 1 − e e ⋅ 1 ⇒ L = 1 − e e ∴ α = e 1 − e e ⇒ log α = 1 − e e ∴ Required value = 1 + 1 − e e 1 − e e = e
Practice Limits on Quantrex Academy →
← Previous question _ x arrow ∞ (2 x^2-3 x+5)(3 x-1)^ (x)/(2) (3 x^2+5 x+4) (3 x+2)^x is equal to :Next question → Let k ∈ ℝ . If _ x arrow 0^+ ( ( k x)+ x+x)^ (2)/(x) =e^6 , then the value of k is
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