Limits, Continuity & Differentiability
Limits
nta_pyq_2025_jan
Grade 12
Question:
If lim x\to\infty (( e )( 1 - x )) = \alpha , then the value of log e \alpha equals : 1-e e 1+x 1+log \alpha e
Step-by-Step Solution
Key Concept: Apply the core result for standard limits and expansions and simplify using the given constraints.
e 1 x \infty \alpha = lim (( )( - )) (1 form ) (4) x\to\infty 1 - e e 1 + x L \therefore \alpha = e e 1 x Where L = lim x (( )( - ) - 1) x\to\infty 1 - e e 1 + x e 1 x 1 - e \Rightarrow L = lim ( )x( - - ( )) x\to\infty 1 - e e 1 + x e e x \Rightarrow L = lim x (1 - ) 1 - e x\to\infty 1 + x e x \Rightarrow L = lim 1 - e x\to\infty x + 1 e \Rightarrow L = ⋅ 1 1 - e e \Rightarrow L = 1 - e e e \therefore \alpha = e 1-e \Rightarrow log \alpha = 1 - e e 1-e \therefore Required value = = e e 1 + 1-e
Correct Answer: 4